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Topology Course Notes — Harvard University — Math 131 Fall 2013 C. McMullen Contents 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 Background in set theory . . . . . . . . . . . . . . . . . . . . . 3 3 Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 4 Connected spaces . . . . . . . . . . . . . . . . . . . . . . . . . 23 5 Compact spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 27 6 Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 7 Normal spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 8 Algebraic topology and homotopy theory . . . . . . . . . . . . 45 9 Categories and paths . . . . . . . . . . . . . . . . . . . . . . . 47 10 Path lifting and covering spaces . . . . . . . . . . . . . . . . . 53 11 Global topology: applications . . . . . . . . . . . . . . . . . . 58 12 Quotients, gluing and simplicial complexes . . . . . . . . . . . 61 13 Galois theory of covering spaces . . . . . . . . . . . . . . . . . 66 14 Free groups and graphs . . . . . . . . . . . . . . . . . . . . . . 75 15 Group presentations, amalgamation and gluing . . . . . . . . . 81 1 Introduction Topology is simply geometry rendered flexible. In geometry and analysis, we have the notion of a metric space, with distances specified between points. But if we wish, for example, to classify surfaces or knots, we want to think of the objects as rubbery. Examples. For a topologist, all triangles are the same, and they are all the same as a circle. For a two–dimensional example, picture a torus with a hole 1.

in it as a surface in R3. This space can be almost completely flattened out. Indeed, it is topologically the same as a rotary with an underpass connecting its inner and outer edges. Figure 1. Rotary with underpass. If we take an ordinary band of paper and put in a twist, it becomes a M¨obius band, which seems clearly different, but how so? What happens if you cut a M¨obius band down the middle? What happens if you do it again? Is a configuration of two linked circles in space fundamentally different from two unlinked circles? The idea of a topological space. The property we want to maintain in a topological space is that of nearness. We will allow shapes to be changed, but without tearing them. This will be codified by open sets. Topology underlies all of analysis, and especially certain large spaces such as the dual of L∞(Z) lead to topologies that cannot be described by metrics. Topological spaces form the broadest regime in which the notion of a continuous function makes sense. We can then formulate classical and basic theorems about continuous functions in a much broader framework. For example, an important theorem in optimization is that any continuous function f : [a, b] → R achieves its minimum at least one point x ∈ [a, b]. This property turns out to depend only on compactness of the interval, and not, for example, on the fact that the interval is finite–dimensional. Invariants. A second agenda in topology is the development of tools to tell topological spaces apart. How is the M¨obius band to be distinguished from the cylinder, or the trefoil not from the figure–eight knot, or indeed how is R3 different from R4? Our introduction to the tools of algebraic topology provides one approach to answer these questions. 2.

This course. This course correspondingly has two parts. Part I is point– set topology, which is concerned with the more analytical and aspects of the theory. Part II is an introduction to algebraic topology, which associates algebraic structures such as groups to topological spaces. We will follow Munkres for the whole course, with some occassional added topics or different perspectives. We will consider topological spaces axiomatically. That is, a topological space will be a set X with some additional structure. Because of the gener- ality of this theory, it is useful to start out with a discussion of set theory itself. Remark on writing proofs. When you hit a home run, you just have to step once on the center of each base as you round the field. You don’t have to circle first base and raise a cloud of dust so the umpire can’t quite see if you touched the base but will probably give you the benefit of the doubt. 2 Background in set theory The axioms of set theory. • Axiom I. (Extension) A set is determined by its elements. That is, if x ∈ A =⇒ x ∈ B and vice-versa, then A = B. • Axiom II. (Specification) If A is a set then is also a set. • Axiom III. (Pairs) If A and B are sets then so is. From this axiom and ∅ = 0, we can now form =, which we call 1; and we can form, which we call 2; but we cannot yet form. • Axiom IV. (Unions) If A is a set, then � A = is also a set. From this axiom and that of pairs we can form � = A∪B. Thus we can define x+ = x+1 = x∪, and form, for example, 7 =. • Axiom V. (Powers) If A is a set, then P(A) = is also a set. 3.

• Axiom VI. (Infinity) There exists a set A such that 0 ∈ A and x + 1 ∈ A whenever x ∈ A. The smallest such set is unique, and we call it N =. • Axiom VII (The Axiom of Choice): For any set A there is a function c : P(A) − → A, such that c(B) ∈ B for all B ⊂ A. Discussion of the Axioms. Axiom I. (Extension). To have a well– defined domain of discourse, the elements of sets are also sets. There is a subtle point in Axiom I: what does the conclusion, A = B, mean anyway? In fact the idea of equality is a notion in logic rather than set theory. It means that for any logical sentence P(x), P(A) has the same answer as P(B). For example, if A = B, and A ∈ Y , then B ∈ Y . Axiom II. (Specification). Examples: A ∩ B =. A − B =. R − Q = irrationals; Q − R = ∅. = even numbers. =. = ∅. For more advanced set theory, one uses the Axiom of Replacement instead of Specification; this permits the construction of cardinals such as ℵω, but it is not required for most ‘mainstream’ mathematics. Assuming at least one set A exists, we can now form 0 = ∅ =, but nothing else for sure. (E.g. A might be ∅.) The Barber of Seville; Russell’s paradox. If X =, is X ∈ X? There is no universe: given a set A, set X =. We claim X ̸∈ A. Indeed, if X ∈ A, then X ∈ X iff X ̸∈ X. One solution to the classic paradox — who shaves the barber of Seville? — is of course that the barber is a woman. In the G¨odel-Bernays theory, you are allowed to form X, but X is not a set; it is called a class. Axiom III. (Pairs). From this axiom and ∅ = 0, we can now form =, which we call 1; and we can form, which we call 2; but we cannot yet form. 4.

Axiom IV. (Unions). From this axiom and that of pairs we can form � = A ∪ B. Thus we can define x+ = x + 1 = x ∪, and form, for example, 7 =. Intersections. If A ̸= ∅, we can define � A =. Since A has at least one element B0, we have � A ⊂ B0 and thus the intersection is a set. Note: � ∅ is undefined! Examples: � = A, � = A ∩ B. Axiom V. (Powers) Examples: X = is the number of possible poker hands. |X| = 2, 598, 960. Pascal’s triangle. The subsets with k + 1 elements of can be partitioned into those that include n and those that do not. Thus �n+1 k � = �n k � + � n k−1 � . Axiom VI. (Infinity) . We have now built up the natural numbers via set theory, and can proceed to the real numbers, functions on them, etc., with everything resting on the empty set. Another standard assumption we have not listed is the Axiom of Exten- sion, which asserts there is no decreasing sequence . . . x3 ∈ x2 ∈ x1. This implies all sets rest on the empty set, and we never have the infinite loop x ∈ x. A kindly mathematician uncle asks his niece, “What’s the highest number you know?” The niece replies, “168,000,000”. The uncle asks, “But what about 168,000,001”? And the niece replies, “I was close, wasn’t I?” (Hubbard) Moving along. We can now define ordered pairs by (a, b) =,}. Then (a, b) = (a′, b′) iff a = a′ and b = b′. Then we can define the product of two sets by A × B =. Note that A × B ⊂ P(P(A ∪ B)), so it is a set. Relations. A relation R between A and B is a subset R ⊂ A × B. It has a domain and range. A relation can be visualized as a directed graph with vertices A ∪ B and with an edge from a to b exactly when (a, b) ∈ R. Examples: an equivalence relation is a subset of A × A with certain properties. The relation i < j on Z. The relation b|a on. 5.

Functions. A function f : A → B is a relation between A and B such that for each a ∈ A, there is a unique b such that (a, b) ∈ f. We write this as b = f(a). Functions are also called maps. The set of all f : A → B is denoted BA. Why? How many elements does 35 have? (Answer: 243.) A function can be injective and/or surjective. It is bijective if both. Composition of maps: f ◦ g. If f : A → B is bijective, then there is a unique map g : B → A such that g ◦ f(x) = x ∀x ∈ A. Examples: f(n) = n2 is injective on N, but not on Z. It is surjective in neither case. The function sin : R → [−1, 1] is surjective but not injective. Its restriction, sin : [−π/2, π/2] → [0, 1], is bijective. Its restriction, sin : [0, 1] → [−1, 1], is injective but not surjective. Set theory as a programming language. The point of the definitions of N and (a, b) is not so much that they are natural or canonical, but that they work. In other words set theory provides a very simple language in which the rest of mathematics can be implemented. There is a natural bijection between A × A and A2. There is a natural bijection between P(A) and 2A. P(X) as an algebra. If we define A · B = A ∩ B and A + B = (A ∪ B) − (A ∩ B), then P(X) becomes a ring. The identity elements are ∅ and X. This rings is nothing but the ring of map f : X → Z/2. Note that A + A = 0. Thus this ring is also an algebra over the field F2. We let χ : P(X) → 2X denote the map that sends A to its indicator function χA which is 1 on A and otherwise zero. Functions, unions and intersections. Let f : X → Y be a function. We set f(A) =. In this way we obtain a map f : P(X) → P(Y ). Is this map a ring homomorphism? In general, f(A∩B) ̸= f(A)∩f(B). We only have f(A∩B) ⊂ f(A)∩f(B). However, if f is injective, then equality holds. We always have, however, f(A ∪ B) = f(A) ∪ f(B). If f : A → B is a function, for any subset X ⊂ B we define f −1(X) =. Thus we have f −1 : P(B) → P(A). This map preserves intersection and unions: e.g. f −1(X ∩ Y ) = f −1(X) ∩ f −1(Y ). Abusing notation, one also writes f −1(b) for f −1(). 6.

The ubiquity of f −1. The fact that f −1 preserves set-theoretic operations means that a good theory of maps often turns on properties of f −1 rather than f. For example, we will see that a continuous function can be defined as one such that f −1(U) is open for every open set U. Similarly a measurable function is one such that f −1(U) is measurable for every open set U. Another reason for this ubiquity is that functions naturally pull back. In particular, the indicator functions satisfy χA ◦ f = f ∗(χA) = χf−1(A). This pullback operation preserves the algebra structure of the space of functions. In differential topology, we find that forms pullback as well, that d(f ∗ω) = f ∗(dω), etc. Cardinality and the Axiom of Choice. We say sets A and B have the same cardinality if there is a bijection between A and B. We will write this relation as |A| = |B|. It is an equivalence relation. Theorem 2.1 |N| ̸= |R|. Proof. Suppose x(n) is a list of all real numbers, and write their fractional parts as = 0.x1(n)x2(n) . . . in base 10. Now choose any sequence of digits yi with yn(n) ̸= xn(n). We can also arrange that yi is non-repeating. Then y = 0.y1y2y3 . . . disagrees with xn in its nth digit, so it is not on the list. Theorem 2.2 (Cantor) . A and P(A) do not have the same cardinality. Proof. Given f : A → P(A), let B =. Suppose B = f(a). Then a ∈ B iff a ̸∈ B. This is a contradiction so f does not exist. 7.

If we think of P(N) ∼= 2N as sequences (ai) of binary digits, then the proof that |N| ̸= |P(N)| is almost the same as digit diagonalization. Corollary 2.3 There are many different sizes of infinity. Finite sets. A set A is finite iff there is a bijection f : A → n for some n ∈ N. That is, A is finite iff |A| = |n| for some n ∈ N. Theorem 2.4 (Pigeon-hole principle) If A is finite, then any injective map f : A → A is surjective. Proof. By induction on |A| = n. Application: for any prime p and a ̸= 0, there is an integer b such that ab = 1 mod p. (I.e. b = 1/a). Proof: The map b �→ ab is 1 − 1 on Z/p, so it is onto. Example: 1/10 = 12 mod 17; in fact 10 ∗ 12 = 120 = 7 ∗ 17 + 1. Infinite sets. A set is infinite iff it is not finite. Theorem 2.5 N is infinite. Proof. Otherwise, there would for some n be an injective map N �→ n, and hence an injective map n + 1 �→ n. This contradicts the pigeon–hole principle. Axiom VII (The Axiom of Choice). For any set A there is a function c : P(A) − → A, such that c(B) ∈ B for all B ⊂ A. In concrete cases it is possible to find explicit choice functions. For the natural numbers, we can let c(A) = min(A). For the rational numbers, we can take this ‘simplest’ rational number x = p/q in A, say with q minimal, |p| minimal given q, and with x ≥ 0 to break ties. The smallest infinite set. Here is typical use of the Axiom of Choice. Theorem 2.6 A is infinite iff there is an injective map f : N → A. Proof. If A is finite then any subset of A is finite, so there is no injection of N into A. Now assume A is infinite; we will construct f. Pick some a ∈ A. Then define, by induction, f(0) = a and f(n + 1) = c(A −). The resulting map is injective by construction. 8.

Cantor’s definition of infinity. A set A is infinite iff there exists a map f : A → A which is injective but not surjective. The proof uses the Axiom of Choice: first embed N into A, and then use x �→ x + 1 on N. Hilbert’s hotel. The set N (or any infinite set) serves to illustrate Hilbert’s hotel. The hotel is full, and yet but just shuffling the residents around we can create an empty room. Note that there is a bijection between N and any infinite subset of N, such that the odd numbers or the squares. There was once a University with a long line of offices had become a little top–heavy: the professors could only occupy the offices with numbers 1, 4, 9, 16, . . . , n2 . . . because the rest were taken up by the Deans. Outraged, the president required that each professor be assigned his own Dean, and the rest fired. The Dean in office n was assigned to the professor in room n2, and now the Dean’s were fully employed as personal assistants to professors. No one had to be fired. In fact the professors in offices 16, 81, 256, ... were still left without assistants, so more Deans were hired. Other applications of AC. Every vector space has a basis. The Hahn– Banach theorem. Every set can be well–ordered. Choice of coset representa- tives for G/H. Existence of non-measurable sets. The Banach-Tarski paradox. As a consequence of AC, you can cut a grapefruit into 5 pieces and reassemble them by rigid motions to form 2 grapefruits. (Now you’ve gone too far.) Relative size. It is natural to say |A| ≥ |B| if there is a surjective map from A to B. But it is equally natural to require that there is an injective map from B to A. The result above is used to show these two definitions are equivalent. Small point: if |B| = 0 then the surjective definition does not work. Let us say |A| ≤ |B| if (1) there is an injection f : A �→ B; or (2) there is a surjection g : B ↠ A, or A = ∅. Theorem 2.7 (1) and (2) are equivalent. Proof. Given the inclusion f we obtain from f −1 a surjection from f(A) back to A, which we can extend to the rest of B as a constant map so long 9.

as A ̸= ∅. Conversely, using the Axiom of Choice, we take f to be a section of g, i.e. set f(a) = c(g−1()). Theorem 2.8 (Schr¨oder-Bernstein) If |A| ≤ |B| and |B| ≤ |A| then |A| = |B|. Proof. We will assume A and B are disjoint — this can always be achieved, if necessary, by replacing A, B with A ×, B ×. Suppose we have injections f : A → B and g : B → A. Then we obtain an injection F = f ∪ g : A ∪ B → A ∪ B. To clarify the proof, say F(x) is the child of x, and x is the parent of F(x). Since F is injective, a child can have only one parent, and every element of A ∪ B is a parent. However some parents are no-one’s child; let us call them godfathers. For any x ∈ A∪B, either x is descended from a unique godfather (possibly x itself), or x has no godfather; it has an infinite line of ancestors (or x is descended from itself.) Now partition A into 3 pieces, A0, AA and AB. A0 is the elements x ∈ A with no godfather; AA consists of those x whose godfather is in A; and AB is those whose godfather is in B. Similarly define B0, BA, BB. There is a bijection A0 ↔ B0 defined by sending a to its child F(a). It is injective because F is, and it is surjective because every x ∈ B0 has a parent, which must lie in A0. There is a bijection AA ↔ BA defined by sending each a ∈ AA to its child F(a). The inverse map sends children to their parents. There are no godfathers in BA, so the inverse is well-defined. Similarly there is a bijection AB ↔ BB, sending a ∈ A to its parent in BB. Putting these three bijections together shows |A| = |B|. Countable sets. We say A is countable if |A| ≤ |N|. Finite sets are countable. Theorem 2.9 If A is countable and infinite, then |A| = |N|. Proof. Infinite implies |N| ≤ |A|, and countable implies |A| ≤ |N|; apply SB. 10.

Theorem 2.10 N2 is countable. Proof. Define a bijection f : N → N2 by f(n) = (a, b) where n + 1 = 2a(2b + 1). Corollary 2.11 A countable union of countable sets is countable. Proof. If X = � Ai we can send the jth element of Ai to (i, j) ∈ N × N. Examples. 1. The set of things that can be described in words is countable. Thus most real numbers have no names. 2. The integers Z can be constructed from 2 × N; they satisfy |Z| = |N|. 3. The rationals Q are Q = Z×Z∗/ ∼, where (a, b) ∼ (c, d) if ad−bc = 0. The Q is countable. 4. The ring of polynomials Z[x] is countable. Corollary 2.12 Most real numbers are transcendental. Warning. While a countable sum of countable sets is countable, the same is not true for products. Indeed, 2N is a countable product of finite sets, but it naturally isomorphic to P(N). Theorem 2.13 |R| = |P(N)|. Proof. We can associate to each subset A ⊂ N a unique real number defined in base 2 by xA = x0.x1x2x3 . . . = � n∈A 2−n, where xn = 1 if n ∈ A and zero otherwise. Conversely, a real number x is uniquely determined by the set Ax = ⊂ Q. Thus |P(N)| ≤ |R| ≤ |P(Q)| = |P(N)|. 11.

Theorem 2.14 |RR| = |P(P(N))|. Thus the functions on R represent the third kind of infinity. Proof. First notice that we have the easy inequality: |P(P(N))| = |P(R)| = |2R| ≤ |RR|. On the other hand, we can construct an injection i : R2 → R by interleaving decimal digits. Since a function f : R → R is a special kind of relation f ⊂ R × R, we then have: |RR| ≤ |P(R2)| = |P(R)|, and so |RR| = |P(P(N))| by the Schr¨oder-Bernstein theorem. Another way to think of the fact that |R2| = |R| is that |N × 2| = |N|, and thus |R2| = |P(N) × P(N)| = |P(N × 2)| = |P(N)| = |R|. The continuum hypothesis. Using the Axiom of Choice, one can prove that for any two sets A and B, |A| ≤ |B| or |B| ≤ |A|. Is there a set A such that |N| < |A| < |R|? It is now known that this question cannot be answered using the axioms of set theory (assuming these axioms are themselves consistent). Some logicians have argued that CH is obviously false (Cohen, Woodin), while others have argued that it must be true (Woodin). 3 Topology This section describes the basic definitions and constructions in topology, together with lots of examples. Topology. A topological space is a set X equipped with a distinguished collection of open sets T ⊂ X. That is, U ⊂ X is open iff U ∈ T . We require that: 1. ∅, X ∈ T ; 2. If U, V ∈ T then U ∩ V ∈ T ; and 12.

3. If A ⊂ T then � A ⊂ T . More informally, any union of open sets is again open, and the intersection of any finite collection of open sets is again open. Usually we will refer to (X, T ) simply as a space (as opposed to a set), and denote it by X. Closed sets. A set F is closed (ferm´e) if its complement U = X − F is open. The axioms above show the collection of closed sets in preserved by arbitrary intersections and finite unions. One can also specify a topology by giving the collection of all closed sets. Euclidean space and metric spaces: Paris and Manhattan. The natural topology T on a metric space (X, d) is defined by saying U is open iff for all x ∈ U there is an r > 0 such that the ball B = B(x, r) satisfies x ∈ B ⊂ U. In other words, there should be some space around each point, like on the wide open range. If you are not so familiar with general metric spaces, you can always think about Rn with d(x, y) = |x − y|. To distinguish metrics and topology, consider the Manhattan metric on Rn: d(0, (x, y)) = |x| + |y|. This gives the same topological space. Exercise. The Paris metric on R2 is defined by d(x, y) is the minimum length of a path from x to y passing through 0, unless x and y happen to lie on the same ray. Exercise: How is the topology of this space essentially different from the usual topology on R2? Bases. This example suggests more generally defining T using a basis B ⊂ P(X). A basis is a collection of sets such that 1. X = � B; and 2. If B1 and B2 in B overlap at x, then there is a B3 ∈ B such that x ∈ B3 ⊂ B1 ∩ B2. We then say U ⊂ X is open if it is a union of elements of B. It is easy to verify that the set T of all such U forms a topology. Closure under unions follows from the definitions, and closure under finite intersections follows from (2). The topology T is simply the smallest topology which contains B. 13.

Clearly the collection B of all open balls B(x, r) form a basis for the usual topology on a metric space. How complex can an open set be? The interval (a, b) form a basis for the topology of R. In fact, every open set can be written as a union of disjoint intervals (exercise). We may need infinitely many intervals to describe U: consider, for ex- ample, U = � N(n, n + 1), � n>0(1/(n + 1), 1/n). Here at least the intervals appear in order. That is not always the case! There exist open sets such that between any two disjoint intervals I, J ⊂ U there is another interval K. A good example is provided by U =. To be more precise, we define U0 = (0.5, 0.6) as the set of numbers that have a robust 5 as their first digit after the decimal point. We have excluded 0.5 and 0.599999 . . ., since these are not robust — a small perturbation gets rid of the 5. Then we define Un = ∈ U0}. Here = x − [x] is the fractional part of x. These are the numbers with a robust 5 in the nth decimal place. Finally we define U = � Un. Then U1 contains the new interval (0.05, 0.06), (0.15, 0.16), (0.25, 0.26), etc. as well as the intervals like (0.55, 0.56) that are already contained in U0. Each Un+1 contains many new intervals like this. The complement of U in [0, 1] is a Cantor set. Continuous functions. A function f : X → Y is continuous if f −1(U) is open for all open sets U ⊂ Y . Intuitively, continuity means: f(x) is close to f(a) whenever x is close enough to a. To say ‘f(x) is close to f(a)’ is to say that f(x) lies in a neighborhood V of f(a). Let U = f −1(V ). Then f(x) is ‘close to f(a)’ iff x ∈ U. When f is continuous, U is a neighborhood of a, and when x is in the neighborhood U of a, it is ‘close enough to a’ that f(x) ∈ V . Theorem 3.1 The composition of two continuous maps is continuous. 14.

Theorem 3.2 A map f : X → Y is continuous iff f −1(B) is open for all element B in a basis for the topology on Y . Proof. For any open set U = � Bi, the preimage f −1(U) = � f −1(Bi) is a union of open sets, hence open. Example. To verify a map f : X → R is continuous, it suffices to verify that f −1(a, b) is open for all (a, b). Homeomorphism. A bijection between spaces, f : X → Y , is a homeo- morphism if both f and f −1 are continuous. This is the same as saying that U is open ⇐⇒ f(U) is open. Homeomorphic spaces are topologically the same. Warning: it is tempting to think that a bijective continuous map is al- ways a homeomorphism. This is not true! For example, let X = ∪. Then the map f : N → X given by f(0) = 0, f(n) = 1/n for n > 0, is continuous and bijective, but X is not homeomorphic to N. We will later see that a continuous bijective map between compact Haus- dorff spaces is always a homeomorphism. The reals and (0, 1). A metric space is bounded if sup d(x, y) < ∞. One of the most basic differences between metric spaces and topological spaces is that the overall size of a space is hard to detect topologically. As an important example we note: The real numbers R are homeomorphic to the open interval (0, 1). A fancy way to see this is to first note that all bounded intervals (a, b) are related by linear maps, and then note that tan : (−π/2, π/2) → R is a homeomorphism. Many other functions will do the trick. Real-valued functions. We let C(X, Y ) denote the set of all continuous map f : X → Y . We let C(X) = C(X, R) denote the algebra of real–valued continuous functions on X. It is easy to show that if f and g are continuous, then so are the sum and product f + g, fg. Thus C(X) is an algebra. It always contains the constant functions R. In a metric space there are always lots of interesting functions in C(X), since f(x) = d(x, x0) is continuous. In a general topological space, there 15.

might be lots or there might be none. One of our tasks is to construct lots of elements of C(X), under suitable hypotheses on X. Functions and topology. If we broaden our test targets beyond R, the space of continuous functions on X uniquely determines its topology. As a simple example, let Z = with the topology where is open but not is not. Then A is open iff χA is continuous. This shows: Theorem 3.3 The topology on X is uniquely determined by the set of con- tinuous functions C(X, Z). Limits. A neighborhood of b ∈ X is just an open set U with b ∈ U. We say an → b if for any neighborhood U of b, we have an ∈ U for all n ≫ 0. Two basic topologies. In the discrete topology, all sets are open, and all functions are continuous, so C(X) = RX. In the trivial topology, only X and ∅ are open, so C(X) = R. Cofinite topology. A slightly more interesting topology is the cofinite topology. In this topology, A ⊂ X is closed iff |A| < ∞ or A = X. If X is finite, then the cofinite topology is just the discrete topology, but otherwise it is different. In the cofinite topology, if an is an infinite sequence, then an → b for all b ∈ X. In particular limits are not unique! Example: The lower–limit topology: Rℓ . An interesting topological space is given by Rℓ, the real numbers with the intervals [a, b) as a basis of open sets. In this topology, an → a iff an converges to a in the traditional sense and an ≥ a for all n ≫ 0. A function f : Rℓ → R is continuous iff f(an) → f(b) whenever an → b from above. Uphill topology. We can think of Rℓ as the uphill topology. It requires huge effort to climb the hill, so points cannot be approached from below. However it is easy to slide down to them. This lack of symmetry is not consistent with a metric, since the latter satisfies d(x, y) = d(y, x). In fact, Rℓ is not metrizable. Comparison of Rℓ with R. Since [a, b) is not generally open in R, the identity map i : R → Rℓ is not continuous. (I.e. Rℓ is not swept out by a path.) On the other hand, i : Rℓ → R is continuous. 16.

Order topology. If X is given a total ordering, then it becomes a topo- logical space Xo by taking the intervals (a, b) = as a basis. The product topology. Let X and Y be topological spaces. The product topology on X × Y is defined by taking as a basis the sets of the form U × V , where U and V are open sets in X and Y . Theorem 3.4 (1) The product topology is the smallest topology such that the projections of X × Y to X and Y are continuous. (2) A function f : Z → X × Y , given by f(z) = (f1(z), f2(z)), is contin- uous iff each of its coordinates f1 and f2 are continuous. Proof. (2) It suffices to check continuity using the basic open sets U × V in X × Y . For these, f −1(U × V ) = f −1 1 (U) ∩ f −1 2 (V ) is an intersection of two open sets, and hence open. Topologies on the square. The product topology on I × I coincides with the usual metric topology. The lexicographical or dictionary order topology gives a weirder space, I2 o, the ordered square. To understand this topology, we must see what a small interval around a point (a, b) looks like. If 0 < b < 1, then × (b′, b′′) is a typical small interval around (a, b). So open vertical segments qualify as open set, and any union of these is also open. This shows any ‘usual’ open subset of the interior of I2 is also open in I2 o. But what is a neighborhood of (a, 1) in I2 o? If 0 < a < 1 then it must con- tain some whole vertical segment a′ × I! In particular, a small neighborhood of (a, 1) in the usual topology is not open in the order topology. Think of this as convergence in a dictionary. Most of the time, if a sequence of words satisfy wn → w, then eventually the first letter of wn agree with that of w. An exception happens, for example, when w = syzyzgy is the last word in the S section. Then any neighborhood of w contains some word beginning with t. See also azygous, Byzantine, Czech, dysuria, Ezra, fylfot, etc. The subspace topology. If A ⊂ X and X is a topological space, the subspace topology is defined by TA =. 17.

This is the unique topology so that if f : Z → A is continuous ⇐⇒ ι ◦ f : A → X is continuous, where ι : A → X is the inclusion. Interval example. Let I = [0, 1] ⊂ R with the subspace topology. Then [0, a) and (b, 1] are open sets in I, even though they are not open in R. Order example. The ordered square I2 o is a subset of the ordered plane R2 o. Is the order topology on the square the same as the subspace topology? Weirdly, the answer is no (see Munkres, Fig. 16.1). Axes example. Let X = R ∪ iR ⊂ C, with the subspace topology. When is an interval (a, b) along the real axis open in X? Only when 0 ̸∈ (a, b)! Topology as a language. We have already seen how to define open and closed sets. Closure. The closure A of a set A ⊂ X is simply the smallest closed set containing A. This exists because the intersection of closed sets is closed. Interior. Similarly, the interior of A, denoted int(A), is the largest open set contained in A. Boundary. The boundary of A is their difference: ∂A = A − int A. A point x lies in the boundary of A iff ever neighborhood of x meets both A and X − A. Density. We say A ⊂ X is dense if A = X. Example: the boundary of the open unit ball B in R2 is S1. The same is true of the closed unit ball. The boundary of the set of points A in the unit ball B which happen to have rational coordinates is the whole ball B. The interior of A is empty, and A is dense in B (in the subspace topology). Limit points. An essential notion in calculus is the idea of limit. But it is usually expressed in terms of sequences. How can it be expressed just using open sets? Answer: we say x is a limit point of A if every neighborhood of x meets A−. In a metric space, this would imply that there is a sequence an ∈ A with an ̸= x for all n and an → x. Theorem 3.5 The closure of A is the union of A and its limit points. Proof. Suppose x ̸∈ A. If x is not a limit point, then there is a neighborhood U of X which does not meet A. Then A∩U = ∅ as well, so x ̸∈ A. Conversely, if x ̸∈ A then U = X − A is an open neighborhood of x disjoint from A, so x is not a limit point. 18.

A point x ∈ A is isolated if there is a neighborhood U of x such that U ∩ A =. Exercise: every point of a closed set A is either a limit point or an isolated point. The set of isolated points in A form an open subset of A, and the set of limit points form a closed subset of A. Examples: derived sets. The set of limit points A′ of a closed set A ⊂ R is called the derived set of A. It is obtained by just throwing out the isolated points. For example, if A = then A′ = and A′′ = ∅. On the other hand, if A = [0, 1] then A′ = A. Can A′ = A if A does not contain an interval? The answer is yes. The Cantor set K ⊂ [0, 1] has K′ = K even though its interior is empty. Derived sets and the invention of ordinals*. Another remarkable fact, which we will not prove, is that a countable closed set A ⊂ [0, 1] always has an isolated point. Thus A ̸= A′. We can then define A0 = A and An+1 = A′ n. It may happen that all these sets are nonempty, and then A∞ = � An ̸= ∅ (by compactness). We can then continue by setting A∞+1 = A′ ∞, and so in. This led Cantor to the invention of ordinals and much of the set theory we have discussed. Obviously A1 = is a countable closed set with A′ 1 nonempty. If we stick together a sequence of copies of A1 accumulating on 0, we obtain a set A2 with A′′ 2 nonempty. Similarly we can inductively construct closed, countable sets An so that A(n) n ̸= ∅. Then we can strict together (A1, A2, . . .) to create a set Aω such that A(ω) ω = ∞ � i=1 A(i) ω ̸= ∅. Continuing in this way, one can construct for any countable ordinal a count- able closed set Aα such that A(α) α ̸= ∅. Cantor was motivated by a problem in Fourier series. A closed set A ⊂ [0, 2π] is a set of uniqueness if the only way a Fourier series � aneint can converge to zero at each point of [0, 2π] − A is if all its coefficients an are zero. Using derivations, Cantor showed any countable closed set is a set of uniqueness. On the other hand, the linear middle ξ Cantor set K(ξ) is sometimes a set of uniqueness, sometimes not — e.g. the case where 1/ξ is a Pisot number is problematic. See Salem, Algebraic numbers and Fourier analysis. Example: sequences are not enough! Let X = R with the topology T where the closed sets are just R and the countable sets A ⊂ R. Take any 19.

uncountable set in R, say B = (0, 1). Then B = R. But there is no sequence an ∈ B with an → 2! In fact, any sequence in B is already closed! Hausdorff spaces. Suppose an → x ∈ X. Can it also be true that an → y in X, with x ̸= y? The answer is yes: in fact, if X has the trivial topology, then every sequence converges to every point, since the only nonempty open set is X. We say X is Hausdorff if any pair of distinct points have disjoint neigh- borhoods. In this case limits are unique! The Hausdorff condition is one of the Trennungsaxioms; it is traditionally denoted T2. The axiom T1 says points are closed. It is easy to see that T2 =⇒ T1. Example. If X is infinite, the cofinite topology on X is not Hausdorff. But it is T1. This type of example comes up in ‘real life’: if X is a variety over C of dimension > 0, then the Zariski topology on X is not Hausdorff. Theorem 3.6 The product of two Hausdorff spaces is again Hausdorff. A subspace of a Hausdorff space is Hausdorff. Continuity in products. It is tempting to think the a function f(x, y) on X ×Y is continuous if y �→ f(x0, y) and x �→ f(x, y0) are continuous for each fixed x0, y0. But this is not the case. The traditional counterexample on R2, F(x, y) = xy/(x2 + y2) with F(0, 0) = 0, is verifiable but mysterious. Here is a better example. Consider instead a map from R2 into S1 = R/2πZ. As a first attempt we take f(x, y) = θ in polar coordinates. But this is not continuous on lines through the origin. To fix that, we observe that 2θ is continuous, since θ jumps by π when it crosses the origin. Now if we set f(0, 0) = 0, then f(x, y) is continuous on horizontal lines. And if we set f(0, 0) = π, it is continuous on vertical lines. To complete the example, we just set f(x, y) = 4θ(x, y) and f(0, 0) = 0. Then f is continuous on horizontal and vertical lines, but its limit is different along lines through (0, 0) with other slopes, so it is not continuous on R2. 20.

For a real-valued function, one can take cos f(x, y) or sin f(x, y). The same construction would work with any function f(x, y) = F(θ), provided F : S1 → R is continuous, F(θ) = 0 for θ = 0, π/2, π, 3π/2, and F(θ) is not identically zero. The ‘traditional’ counterexample is proportional to f(x, y) = sin(2θ) = Im z2 |z|2 = 2xy x2 + y2· More generally, if we take any two homogeneous polynomials in x, y of the same degree, P and Q, with Q(x, y) ̸= 0 for (x, y) ̸= 0, then we can set f(x, y) = P(x, y) Q(x, y) away from the origin. Then, if P(1, 0) = P(0, 1) = 0, we can extend f by f(0, 0) = 0 to get a function continuous in x and y individually. Finally, if P(x, y) is not identically zero, f(x, y) will not be continuous. Normality. Once points are closed, we can impose the stronger normality condition (T4) that any two disjoint closed sets, A, B ⊂ X, can be engulfed by disjoint open sets, U ⊃ A, V ⊃ B. We will later see that this powerful condition insures there is a continuous function f : X → R such that f|A = 1 and f|B = 0. In particular, C(X) is nontrivial (so long as |X| > 1). Infinite products. Let Xi be a collection of topological spaces indexed by i ∈ I. Their product X = � Xi consists of maps x : I → � Xi such that xi ∈ Xi. The product topology on X is defined by the basis of open sets U = � Ui where each Ui ⊂ Xi is open, and Ui = Xi for all but finitely many i. (Check that these basis elements satisfy B1 ∩ B2 ⊃ B3.) Theorem 3.7 A map f : Z → � Xi is continuous iff each of its coordinates fi : Z → Xi is continuous. Example. Let 2 = be given the discrete topology. Then 2N ∼= P(N) becomes a topological space. Two subsets of N are close if they have the same intersection with [0, N] for large N. 21.

The traditional Cantor set K ⊂ [0, 1] is homeomorphic to 2N, and thus provides a way of visualizing all the subsets of N. It turns out that for any sequence of nonempty finite sets Xi, each with the discrete topology, the space � Xi is homeomorphic to the Cantor set. This is already tricky to see for 3N. Pointwise convergence. Fix any set S, and let X = RS = with the product topology. Then a sequence fn ∈ X converges to g iff fn(s) → g(s) for all s ∈ S. Thus the product topology is sometimes called, in the case X = AB, the topology of pointwise convergence. The Hilbert cube. Another nice example of an infinite product is X = [0, 1]N. This is simply the space of all sequences (xi) with 0 ≤ xi ≤ 1. The topology is induced by a nice metric, e.g. d(x, y) = max |xn − yn|/2n. Box topology. The box topology on X = � Xi is defined by using all products of open sets, � Ui, to form a base for the topology. This topology is very strong (i.e. it has lots of open sets), so there are very few continuous maps into the product. Exercise. The diagonal map D of [0, 1] into [0, 1]N is continuous in the product topology, but discontinuous in the box topology. (Proof: Each factor Di : [0, 1] → [0, 1] is simply the identity map, which is continuous, so the product topology is fine. But in the weak topology, for any x ∈ [0, 1] we can find a sequence of open sets Ui containing x and nesting down to it. Then D−1(� Ui) = � Ui = is not open, so D is not continuous.) Topological groups. A topological group is a set G equipped with contin- uous product and inverse maps, G × G → G and G → G, that satisfy the usual group axioms. The most basic topological groups are (R, +) and (R∗, ·). The fact that these are groups shows that C(X) is an algebra. The fact that inversion is continuous shows that 1/f ∈ C(X) whenever f has no zeros. Other examples include the matrix group GLn(R), which can be regarded via matrices as an open subset of Rn2. Cramer’s rule shows inverse is con- tinuous. Here is a typical theorem about topological groups. 22.

Theorem 3.8 If the origin is closed in G, then G is Hausdorff. Proof. Since right multiplication (say) is a homeomorphism, it suffices to show that identity element e and any other element x ̸= e ∈ G have disjoint neighborhoods. Since e is closed, so is x, so there is a neighborhood U of e disjoint from x. Now comes the key point: since multiplication is continuous, there is a smaller neighborhood V of U such that V · V ⊂ U. We may also assume V = V −1. It is then readily verified that V and xV are disjoint. 4 Connected spaces A space X is connected if it cannot be written as the disjoint union of two nonempty open sets, X = U ∪ V . When X breaks up into two open pieces U and V , each piece is a clopen — it is both open and closed. A subset A ⊂ X is connected if it is connected in the subspace topology. A space is totally disconnected if its only connected subsets are single points. Two basic facts about connectedness. One is concrete and one is ab- stract. The concrete fact is: Theorem 4.1 The interval [0, 1] is connected. Proof. Suppose [0, 1] is the union of two disjoint open sets, U and V . We may assume 0 ∈ U. Let a = sup. Since U is open, a > 0. If a = 1 we are done. Otherwise a ̸∈ U, and therefore a ∈ V . But then a neighborhood of a is in V , so we cannot have [0, a) ⊂ U. The abstract fact is: Theorem 4.2 If f : X → Y is continuous and X is connected, then f(X) is connected. Proof. Suppose to the contrary we can write f(X) = U ∪ V as the union of two nonempty, disjoint open sets. Then X = f −1(U) ∪ f −1(V ) does the same for X. 23.

Exercise. What does this have to do with the intermediate value theorem? Idempotents. It is a beautiful fact that connectedness is detected by idem- potents in the algebra C(K), i.e. elements such that f 2 = f. On a connected space, the only idempotents are f = 0 and f = 1. But if X = U ⊔ V , then χU is continuous and χ2 U = χU. Locally constant functions. By the same token, if X is connected and f : X → R is continuous and locally constant, then f is constant. Path-connected spaces. To show X is connected, it is often easier to draw paths between points than to analyze open partitions of X. We say X is path connected if for all x, y ∈ X there is a continuous map f : [a, b] → X such that f(a) = x and f(b) = y. Theorem 4.3 A path connected space is connected. Proof. Suppose X = U ⊔ V and x ∈ U. Pick a path f : [a, b] → X connecting x to any other point y. Then f([a, b]) is connected, so y ∈ U. Thus V = ∅. The general principle at play here is: if any two points x, y ∈ X lie in a connected subset A ⊂ X, then X is connected. The role of A is played by f([a, b]) in the proof above. Convex sets. A subset K ⊂ Rn is convex if whenever x, y ∈ K, the straight line from x to y also lies in K. Since this straight line is a path, we find: Any convex set K is path connected. Note that we do not require K to be open or closed. Exercise. A set A ⊂ R is connected ⇐⇒ it is convex ⇐⇒ it is path connected. The topologist’s sine curve. The union of the graph of y = sin(1/x) over (0, 1] and the interval [−1, 1] along the y-axis is a classic example of a closed subset of R2 that is connected but not path connected. 0.2 0.4 0.6 0.8 1.0 �1.0 �0.5 0.5 1.0 24.

A somewhat similar set X ⊂ R2 is obtained by taking the union of the graphs of y = 0 and y = 1/x over [0, ∞). But this X is disconnected. What is the difference between these two examples? Path connectedness in Rn. On the other hand, for well-behaved spaces like manifolds, connectedness and path–connectedness are the same. The proof is a typical use of the property of connectedness: we show the set of good points is both open and closed, so it must be the whole space. Theorem 4.4 An open set U ⊂ Rn is connected ⇐⇒ it is path connected. Proof. Suppose U is connected and x ∈ U. Let V ⊂ U be the set of points that can be reached from x by a polygonal path. Clearly V is open. We claim it is also closed (in U). Indeed, if y ∈ V then there is a ball B with y ∈ B ⊂ U, and a z ∈ B that can be reach from x by a polygonal path. But B is convex, so y can also be reached. The converse is a general fact. This proof is much easier than, say, drawing a line between x and y in U, and then explicitly modifying it so it stays in U. Example: Countable metric spaces. A space is totally disconnected if the only connected sets it contains are single points. Theorem 4.5 Every countable metric space X is totally disconnected. Proof. Given x ∈ X, the set D = is countable; thus there exist rn → 0 with rn ̸∈ D. Then B(x, rn) is both open and closed, since the sphere of radius rn about x is empty. Thus the largest connected set containing x is x itself. The continuum. The real numbers R are sometimes called the continuum, because they form a connected set, corresponding to the geometric notion of a line. Points Have no parts or joints How then can they combine To form a line? —J. A. Lindon. 25.

In fact if we only consider points with names, these form a thin (countable subset) sprinkled along the continuum; the uncountably many remaining (fictional) points serve to connect them. A countable connected Hausdorff space. (Due to Urysohn?) It is thus very curious that there are countable connected spaces which are Hausdorff! Of course they are not metrizable or path connected. Let H be the set of points in the closed upper half-plane with rational coordinates. Any point in H is the vertex of a (possibly degenerate) equilat- eral triangle T(p) resting on the x-axis. Let a neighborhood of p consist of p itself union a pair of intervals on the x-axis about the feet of T(p). Since a line with slope 60◦ passes through at most one rational point, the feet of T(p) and T(q) are disjoint if p ̸= q and thus H is Hausdorff. On the other hand, any open set contains an interval on the x-axis, and the closure of such an interval consists of two bands forming an infinite copy of the letter V . Any two of these V ’s intersect, so H is connected. Fans and products. Here are two more basic properties of connectedness. Theorem 4.6 A fan of connected sets is connected. That is, if Ai ⊂ X are connected subspaces which all have a point in common, then Y = � Ai is connected. Proof. Let Y = U ⊔V and suppose the common point p lies in U. Note that U ∩ Ai and V ∩ Ai are disjoint open sets in Ai. Since Ai is connected, and p ∈ U, we have Ai ⊂ U. Thus Y = � Ai ⊂ U and therefore Y is connected. Corollary 4.7 A product of connected sets A × B is connected. Proof. Suppose A×B = U ⊔V , (a, b) ∈ U, and we are given (a′, b′) ∈ A×B. Then A × is connected, so (a′, b) ∈ U. Similarly × B is connected, so (a′, b′) ∈ U. Thus V = ∅. It follows that any finite product of connected sets is connected. Infinite products. Exercise: Show that an infinite product of connected sets is connected. 26.

Components. Because fans are connected, every point x ∈ X lies in a unique maximal connected set A(x). This is called the component of X containing x. Now it is easy to show that the closure of a connected set is connected, so A(x) is closed (else it would not be maximal). In general A(x) is not open; for example, in the Cantor set A(x) = for all x. The components are open when X has only finitely many components. 5 Compact spaces In advanced calculus one learns that the interval [a, b], and more generally any closed, bounded subset of Rn, is compact. Compact spaces K have many important properties. For example, any continuous map f : K → R achieves its maximum. Compact spaces are also ‘bounded’ in a strong sense, and thus they capture some of metric idea of a space of bounded diameter. Unlike the definition of connectedness, the notion of compactness is subtle and requires some practice to use. It is almost always better to work with compact spaces that are also Hausdorff, and indeed Bourbaki has redefined the word ‘compact’ to include this condition. Finite covers. Let X be a topological space. A collection of open sets (Ui) with � Ui = X is an open cover of X. If we can extract from this covering a finite collection of open sets (V1, . . . , Vk) = (Ui1, . . . , Uik) that still cover X, we say (Ui) has a finite subcover. The space X is compact if every open cover of X has a finite subcover. A more succinct statement is: if U is a collection of open sets and X = � U, then there is a finite set V ⊂ U such that X = � V. Examples. The real numbers are not compact, since the covering R = � Z(n, n + 2) has no finite subcover. The interval (0, 1] is noncompact as well: �(1/n, 1] has no finite subcover. But [0, 1] is compact. For example, if we write [0, 1] = [0, ϵ) ∪ � n (1/n, 1], then 1/N < ϵ for some N, and hence we have a finite subcover (by just two elements!) [0, 1] = [0, ϵ) ∪ (1/N, 1]. 27.

Two main results. As for connectedness, there are 2 basic results, one abstract, one concrete. Theorem 5.1 The continuous image of a compact space is compact. Proof. Suppose A is compact and f : A → X is continuous. Let B = f(A) and let � Ui be an open cover of B. Then � f −1(Ui) is an open cover of A. Since A is compact, the covering f −1(Ui) has a finite subcover, say (f −1(Uj) : j ∈ J), |J| < ∞. Since Uj = f(f −1(Uj)), the sets (Uj : j ∈ J) cover B. In analysis one frequently discusses compactness in terms of sequences, but this does not suffice for general topological spaces. Thus it is interesting to prove compactness on the real line directly using the covering condition. Theorem 5.2 The interval [0, 1] is compact. Proof. Let (Ui) be an open cover of [0, 1]. Let A =. Clearly 0 ∈ A. We will show A is open and closed. Then, since [0, 1] is connected, we have will have 1 ∈ A, which is the Theorem. In fact it is obvious that A is open, since any cover that works for [0, x] has x ∈ Ui for some i; hence B(x, r) ⊂ Ui for some r > 0, and hence the same cover works for [0, x + r/2]. The proof that A is closed is similar. Suppose x is a limit point of A. Of course x ∈ B(x, r) ⊂ Ui for some i. Moreover [0, x − r/2] admits a finite cover. Now add Ui to this cover, and we have a finite cover of [0, x]. Of course the same result holds for any other interval. We will shortly see the following basic result: Theorem 5.3 A set K ⊂ Rn is compact iff it is closed and bounded. Compact Hausdorff spaces. We now explain a few points about compact sets versus closed sets. Theorem 5.4 If X is compact, then any closed set A ⊂ X is also compact. Proof. Given any open cover of A, we can use U = X − A to extend it to any open cover of X. We then pass to a finite cover, discard U, and obtain a finite open cover of A. 28.

Here is a subtle point, whose proof is a beautiful application of compact- ness. Theorem 5.5 If X is a Hausdorff space, then any compact set K ⊂ X is closed. Proof. Suppose x ∈ X − K. By the Hausdorff condition, for every y ∈ K there is are disjoint neighborhoods Uy and Vy of x and y respectively. Then K is covered by � Vy. Pass to a finite subcover, say V1, . . . , Vk with corresponding neighborhood U1, . . . , Uk of x. Since a finite intersection of open sets is open, U = U1 ∩ · · · Uk is a neighborhood of x as well. Now this neighborhood is disjoint from K ⊂ V1 ∪ · · · Vk and hence x is not a limit point of K. Thus K is already closed. Cofinite revisited. Let X be an infinite set (say R) with the cofinite topology. Then X is compact — as is any subset of X. But there are lots of subsets of X that are not closed (any infinite set A ̸= X). Continuity and compactness. One great feature of compact spaces is that proofs that maps are homeomorphisms are twice as easy. Corollary 5.6 A continuous bijection f : X → Y between compact Haus- dorff spaces is a homeomorphism. Proof. It suffices to show that whenever A ⊂ X is closed, so is f(A). But if A is closed, it is compact; thus f(A) is also compact; and since Y is Hausdorff, this means f(A) is closed as well. This result is certainly false without compactness. The simplest coun- terexample is the natural map f : [0, 1) → S1 = R/Z. Theorem 5.7 A product of compact sets is compact. Proof. An open covering of A × B can be refined so its elements are all of the form U ×V , with U and V open. It suffices to extract a finite cover from this refinement. 29.

Now for any a ∈ A, the set × B is compact, so it has a finite cover of the form �n 1 Ui × Vi. We may assume a ∈ Ui for all i. Then W = � Ui is a neighborhood of a, and W × B admits a finite cover. Since A is compact, can cover it by finitely many open sets Wi of this form. Each product Wi×B admits a finite cover, so their union A × B admits a finite cover as well. Corollary 5.8 A set F ⊂ Rn is compact iff it is closed and bounded. A stronger result is due to Tychonoff: Theorem 5.9 Given any collection of compact spaces Xi, the space � Xi is compact in the product topology. Here is a curious remark: if the Xi are nonempty, how do we even know that � Xi is nonempty? To show this in general requires the Axiom of Choice! We will prove Tychonoff’s theorems for countable products of metric spaces later. The general case can be handled similarly, using nets and transfinite induction, although modern proofs are much more slick. Compactness and closed sets. Let Fn ⊂ R be a sequence of nonempty closed sets with F1 ⊃ F2 ⊃ · · · . Can it be that � Fn is empty, even though at every finite stage it is nonempty? The answer is yes: e.g. take Fn = [n, ∞). But the answer is no for Fn ⊂ [0, 1]. More generally we have: Theorem 5.10 Suppose X is compact. Let F ⊂ P(X) be a collection of closed sets such that F1 ∩ · · · Fk ̸= ∅ for all finite subsets ⊂ F. Then � F ̸= ∅. Proof. If � F = ∅ then � U = X, where U =. Then U has a finite subcover, with U1 ∪ · · · ∪ Uk = X. Taking complements, we obtain elements of F such that F1 ∩ · · · ∩ Fk = ∅. Example. From the ratios fn+1/fn of the Fibonacci numbers (1, 1, 2, 3, 5, 8, 13, 21, . . .) we obtain a nested sequence of intervals (Fn) = ([1, 2], [3/2, 2], [3/2, 5/3], [8/5, 5/3], . . .). 30.

We have � Fn =. Proof of Theorem 5.3. If K ⊂ Rn, then it is closed because Rn is Haus- dorff, and it is bounded because � B(0, n) has a finite subcover. Conversely, if K is bounded then it is contained in a product of intervals [−M, M]n, which is compact. Thus if K is also closed, it is a closed subset of a compact space, and hence compact. Calculus. The following typical applications are crucial in science, eco- nomics, etc. Corollary 5.11 A continuous function f : X → R on a compact space achieves its maximum. Proof. In this case f(X) ⊂ R is compact, so it is closed and bounded. Thus m = sup f(X) < ∞ and m ∈ f(X), so m = f(x) for some x ∈ X. Corollary 5.12 Let f : K → R be a smooth function on a closed, bounded domain in Rn. Then f achieves its maximum at a point x ∈ ∂K, or at a point x ∈ int(K) where Df = 0. Nets*. Sequences have a beautiful generalization called nets that restores their usefulness for general topological spaces. A directed set I is a partially ordered set such that for any i, j ∈ I, there is a k ∈ I with i < k and j < k. A net is a map x : I → X from a directed set into a topological space. We say xi → y if for every open neighborhood U of y, there is an n ∈ I such that xi ∈ U for all i > n. Theorem 5.13 If x is a limit point of A ⊂ X, then there is a net ai ∈ A that converges to x. Proof. Let I be the set of all neighborhoods of A, with U < V if V ⊂ U. Since x is a limit point of A, by the Axiom of Choice there is a net a : I → A such that aU ∈ U ∩ A. It is then a tautology that aU → x. 31.

A subset E ⊂ I is cofinal if for every i ∈ I there is a j ∈ E with j > i. If J is another directed set, and π : J → I is an order–preserving map such that π(J) is cofinal in I, we say xπ(j) is a subnet of xi. Theorem 5.14 A space X is compact iff every net (xi) in X has a conver- gent subnet. Proof. Suppose every net has a convergent subnet. Let F be a family of closed sets with the finite intersection property. Choose a point xA ∈ � A for each finite subset A ⊂ F. Pass to a subnet so xi → x ∈ X. Now for any particular F ∈ A, we have F ∈ A for all A >. Thus xi ∈ F for all i ‘sufficiently large’, and hence x ∈ F. It follows that x ∈ � F and hence X is compact. The converse is similar. With this terminology in place, the product theorem for a pair of general compact spaces can be proved by passing twice to subnets as in the usual argument for metric spaces. Local and global perspective. Here are a few formal similarities between the properties of compactness and connectedness: • Both are preserved by continuous maps, A �→ f(A); • Both are preserved by products, A × B; and • The interval [0, 1] is both compact and connected. However compactness has the stronger hereditary property that if X is com- pact and A ⊂ X is closed, then A is also compact. This means that once the ambient space X is compact, compactness becomes a local property, while connectedness remains a global property. 6 Metric spaces In this section we discuss the topology of metric spaces. Especially we show the metric space C(K) is complete when K is compact, and discuss its com- pact subsets (these are spaces of functions). 32.

Sequences. Let (X, d) be a metric space, given its usual topology: a basis for which consists of all the open balls B(x, r) in X. Note that X is Hausdorff, so limits are unique. Recall that xi → x iff d(xi, x) → 0. One of the main differences between X and a general topological space is that we can think purely in terms of sequences. Thus: 1. The closure of A ⊂ X is given by taking all limits of convergence sequences with xi ∈ A. 2. A function f : X → Y between metric space is continuous iff f(xi) → f(x) whenever xi → x. 3. The space X is compact iff every sequence has a convergent subse- quence. Completeness. A Cauchy sequence in a metric space is a sequence xi ∈ X such that lim N→∞ sup i,j>N d(xi, xj) = 0. We say X is complete if every Cauchy sequences converges. This means whenever d(xi, xj) → 0, we have xi → x for some x ∈ X. A Cauchy sequence is like a swarm of bees. In a complete space, you can see what they are after. Any metric space can be naturally embedded in a complete metric space. The real numbers R are the completion of Q; this is their basic raison d’ˆetre. (Paradox: it is tempting to define R as the completion of Q, but then we need to first define complete metric spaces — without reference to R!) One can also complete the rational numbers with respect to the m-adic valuation, where |mkq| = m−k for integers with gcd(m, q) < m. For example, Z10 consists of decimal numbers which are infinite to the left. It is homeo- morphic to XN, where X =. For p prime, the completion Qp is a field, but Q10 is not. If 5n accumulates on x and 2n accumulates on y, then |x|10 = |y|10 = 1 but xy = 0. More analytical examples of a complete metric spaces come from mea- surable functions, such as L2[0, 1]. Indeed, a major program in analysis is to study completions of spaces of functions with respect to various norms or metrics. The space L2[0, 1] is the completion of C[0, 1] with respect to the norm ∥f − g∥2 = � 1 0 |f(x) − g(x)|2 dx. 33.

The theory of Lebesgue measurable functions is design to give concrete rep- resentatives for elements of this space, just as the real numbers describe elements of the completion of Q. Sequential compactness. Compactness has an alternative intuitive defi- nition in metric spaces. We say X is sequentially compact if every sequence has a convergent subsequence. Theorem 6.1 For metric spaces, compactness and sequential compactness are equivalent. For the proof is it useful to first observe: Theorem 6.2 A sequentially compact metric space X has a countable base. Proof. It suffices to show that for each n, we can find a finite set of open balls Bn of radius 1/n that cover X. To see this, choose a maximal E ⊂ X with d(e, e′) ≥ 1/n for all e ̸= e′ in E. Then E is finite, since an infinite se- quence of distinct points in E would have no limit point, violating sequential compactness. We claim the balls B(x, 1/n) with x in E cover X. Indeed, if some point y ∈ X were left over, then we would have d(y, x) ≥ 1/n for all x ∈ E, so we could add y to E. It is now easy to check that B = � Bn gives a countable base for X. Proof of Theorem 6.1. Suppose X is compact and xn is a sequence in X. We may assume the elements of the sequence are all distinct. Let Fn be the closure of the set. Then E = � Fn is nonempty by compactness. If y ∈ E than for any n > 0 we can find an xin ∈ B(y, 1/n). Thus we have a subsequence converging to y. The converse is a little trickier. Suppose X is sequentially compact, and U is an open cover of X. As we have seen, X has a countable base. Thus we can assume U is a countable cover of X, by choosing for each basis element B a single element of U that contains it (if one exists). We can now order the open sets as U1, U2, . . ., and we want to show that Vn = �n 1 Ui is equal to X for some n. Equivalently, setting Fn = X − Vn, it suffices to show that if we have a sequence of nonempty closed sets F1 ⊃ F2 ⊃ · · · , then F = � Fi ̸= ∅. This is easy: just choose xi ∈ X, and use sequential compactness to obtain a subsequence converging to y; then y ∈ F. 34.

Corollary 6.3 A compact metric space is complete. Proof. Any Cauchy sequence xi has a subsequence with a limit y; but then xi → y as well. Lebesgue number and uniform continuity. Here are 2 useful facts about functions and coverings in metric spaces. Theorem 6.4 Let U be an open cover of a compact metric space K. Then there exists an r > 0 such that for any x ∈ K, we can find U ∈ U such that B(x, r) ⊂ U. Proof. Let Vs ⊂ K be the set of x such that B(x, s) is contained in some element of U. Clearly � s>0 Vs = K. Since K is compact, we have K = Vr for some r > 0. and then r = min(s1, . . . , si) works. Corollary 6.5 Let f : K → R be continuous. Then for any ϵ > 0 there exists an r > 0 such that d(x, y) < r =⇒ |f(x) − f(y)| < ϵ. Proof. Let U be the covering of K by the preimages under f of all intervals of length ϵ in R, and apply the preceding result. Uniform limits. Let Y X denote the set of all functions f : X → Y , and let C(X, Y ) ⊂ Y X denote the subset of continuous functions. Suppose Y is a metric space. Then we can define a distance between functions by d(f, g) = sup X |f(x) − g(x)|. This is not quite a metric, since it may be infinite. We can rectify this by simply setting d(f, g) = 1 if the supremum above is > 1. We say fn → g uniformly if d(fn, g) → 0. We call d the uniform metric. Theorem 6.6 If Y is complete, then so is (Y X, d). Proof. Suppose fn ∈ Y X is a Cauchy sequence. Then so is fn(x); thus fn(x) → g(x), and d(fn, g) ≤ sup m>n d(fn, fm) → 0 since (fn) is a Cauchy sequence. 35.

Theorem 6.7 A uniform limit of continuous functions is continuous. In other words, C(X, Y ) is closed in Y X. Proof. Suppose fn ∈ C(X, Y ) converge uniformly to g : X → Y . Let V ⊂ Y be open. We must show that U = g−1(V ) is open. Suppose x ∈ g−1(V ), and y = g(x) ∈ V . Then B(y, r) ⊂ V for some r > 0. Let Un = f −1 n (B(y, r/2)). Clearly Un is open for all n. Choose N so d(fn, g) < r/2 for all n > N. Then if d(fn(x), y) < r/2, we have d(g(x), y) < r, and hence Un = f −1 n (B(y, r/2)) ⊂ g−1(B(y, r)) ⊂ U for all n > N. Moreover fn(x) → g(x) = y, so x ∈ Un for all n ≫ 0. It follows that x ∈ Un ⊂ U for all n ≫ 0, and thus U is open. Corollary 6.8 If Y is complete, then so is C(X, Y ). Computer programs. Suppose you are asked to write a computer program that takes as input x ∈ [0, 1] and outputs F(x) = 1 if x = 1 and F(x) = 0 otherwise. Unfortunately you have not yet learned how to use branching in your program. So you propose a program that just uses iteration: it computes fn(x) = xn, and outputs F(x) = lim fn(x). Now of course, in practice you can only compute finitely many steps. And even if x < 1, you won’t get fn(x) < 1/2 until you have iterated about n = |1/ log2(x)| times. More and more iterations are required, the closer x is to one! This nonuniformity results from the fact that you are trying to compute a discontinuous function F(x), without using if statements. So even though fn → F pointwise, it cannot do so uniformly — otherwise F would be continuous. Figure 2. Peano curve motif. 36.

Peano curves. An amusing application of completeness is the construction of a Peano curve, that is a continuous, surjective map f : I → I2. This shows that the dimension of a set can go up when we take its continuous image! Cantor was already amazed when he saw that there is a set-theoretic bijection between I and I2. (The basic idea of this is easy: send 0.x1x2x3 . . . to (0.x1x3x5 . . . , 0.x2x4x6 . . .).) More interesting is the fact that a surjective map I → I2 can be made continuous. Exercise: there is no continuous bijection f : I → I2. (Use connectivity.) To construct a Peano curve it suffices to show: Theorem 6.9 There exists a uniformly convergent sequence of functions fn : I → I2 such that fn(I) enters every dyadic square of side 2−n. One such construction (there are many!) is shown in Figure 2. Whenever we have a segment J such that one endpoint of f(J) passes through the center of a dyadic square S, we can modify f|J so that it also passes through the centers of the 4 adjacent subsquares. In doing so we never move f(x) a distance greater than diam(S) = 2−n. Since � 2−n is finite, the sequence converges uniformly. The sequence is shown in Figure 3. Products of metric spaces. Using sequential compactness, it is quite easy to see that the product of two compact metric spaces is compact. Namely, if (xi, yi) is a sequence in X × Y , we can pass first to a subsequence such that xi → x, and then to a subsequence such that yi → y. Then (xi, yi) → (x, y). By diagonalization, it is similarly easy to see that a countable product of compact metric spaces, �∞ 1 Xi, is compact. We can now give a ‘new’ proof that [0, 1] is compact: the product 2N is obviously compact, and there is a surjective, continuous function f : 2N → [0, 1] given by f(a0, a1, a2, . . .) = � 2−i−1ai. Characterization of compact metric spaces. As we have seen, K ⊂ Rn is compact iff it is closed and bounded. Here is a generalization. We say a metric space X is totally bounded if for every r > 0, there is a covering of X by finitely many balls of radius r. Theorem 6.10 A metric space is compact iff it is complete and totally bounded. Proof. We have already seen that a compact metric space is complete; and total boundedness follows by taking a finite subcover of the covering by all balls of radius r. 37.

Figure 3. Approximate Peano curves. 38.

Now suppose X is complete and totally bounded. Let xn be a sequence in X. Given r > 0, we can cover X by finitely many r balls. Then there is a subsequence that lies entirely in one of these balls; it satisfies d(xi, xj) < 2r for all i, j. Diagonalizing, we obtain a subsequence of (xn) which is also a Cauchy sequence. By completeness, this subsequence has a limit y, and hence X is (sequentially) compact. Functions on a compact space. Whenever K is compact, we can make C(K) = C(K, R) into a complete metric space by setting d(f, g) = ∥f − g∥ = sup X |f(x) − g(x)|. This supremum is finite because X is compact. The Arzela–Ascoli theorem. As a vector space, C[0, 1] is infinite–dimensional. Unlike Rn, its closed, bounded subsets are not compact. For example, fn(x) = xn has no convergent subsequence; neither does fn(x) = sin(nx). One of goals is to describe the compact subsets of C[0, 1] and more gen- erally of C(X), when X is a metric space. Here is a typical application of the preceding fact. Theorem 6.11 Let F ⊂ C([0, 1]) be the set of differentiable functions with |f(x)| ≤ 1 and |f ′(x)| ≤ 1. Then F is compact. In other words, a sequence of bounded functions with bounded derivatives has a uniformly convergent subsequence. Proof. To understand this result, it is important to first see that the unit ball in C([0, 1]), unlike the unit ball in Rn, is not compact. (Consider the sequence fn(x) = xn.) But in this example, f ′ n(1) = n → ∞. We cannot make an example like this if we keep |f ′| bounded! The reason for this is that F (or equivalently F) is totally bounded. Here is a proof. Let us say f, g ∈ C([0, 1]) are close along the 1/n grid if |f(a/n) − g(a/n)| < 1/n for a = 0, 1, . . . , n. It is easy to find a finite set Gn ⊂ F such that every f ∈ F is close to some gn ∈ Gn along the 1/n grid. But then, by the intermediate value theorem, we also have |f(x) − gn(x)| ≤ 2/n 39.

for all x. That is, the balls B(gn, 2/n) cover F and the balls B(gn, 3/n) cover F. Thus F is totally bounded. Since C(K) is complete, so is F. Thus F is compact. Equicontinuity. More generally, if K is a compact metric space and F ⊂ C(K) is a family of continuous functions on K, we say F is equicontinuous if for all ϵ > 0 there exists a δ > 0 such that d(x, y) < δ =⇒ |f(x) − f(y)| < ϵ for all f ∈ F. (In the example above, we can take δ = ϵ.) This property allows us to replace ‘close on a grid’ with ‘close throughout K’ as we did in the proof above. Thus the same argument shows: Theorem 6.12 (Arzela–Ascoli) Let K be a compact metric space, and let F ⊂ C(K) be an equicontinuous family of uniformly bounded functions. Then F is compact. Complex analysis and normal families. This result is not so powerful in calculus but it is very powerful in complex analysis. For example, one can use it to show that any sequence of analytic functions fn : U → C, satisfying |fn(z)| ≤ M for all n and z ∈ U, has a subsequence that converges uniformly on compact subsets of U to another analytic function g(z). This is because, by Cauchy’s integral formula, we have |f ′(z)| = ���� 1 2πi � γ f(ζ) (z − ζ)2 dζ ���� ≤ M d(z, ∂U)· Baire category. Here is a final remark about complete metric spaces. If X is compact, then any nested intersection � Fi of nonempty closed sets is nonempty. Theorem 6.13 (Baire category) If X is complete, then the intersection �∞ 1 Ui of any sequence of dense open sets is nonempty (and dense). Proof. By density we can choose a nested sequences of closed balls B(xi, ri) ⊂ Ui with ri → 0. Then (xi) forms a Cauchy sequence, and y = lim xi lies in B(xi, ri) so it lies in � Ui. This shows � Ui is nonempty; and by varying the choice of B(x1, r1), we see it is dense. 40.

Corollary 6.14 If X is a complete metric space and F1 ⊂ F2 ⊂ · · · is an increasing sequence of closed sets with X = � Fi, then one of them has nonempty interior. Example. (A challenging homework problem.) Find a sequence of open sets Ui ⊂ R such that � Ui = Q. Answer: this is impossible! Let Q = and let Vi = Ui −. Since Ui contains Q, each Vi is a dense open set; but if � Ui = Q then � Vi = ∅. Baire category is frequently useful in the construction of counterexam- ples. For example, rather than showing there exists a nowhere differentiable function f ∈ C[0, 1], one can show that a generic function is nowhere differ- entiable. Example: Liouville numbers. We say x ∈ R is a Liouville number if x is irrational and, for any k, there are infinitely many rational numbers with ����x − p q ���� ≤ 1 qk · In other words, x is very well approximated by rational numbers. For exam- ple, x = �∞ 0 1/10n! is a Liouville number, since xm = �m 0 1/10n! = pm/qm with qm = 10m!, and |x − pm/qm| ≤ 2 · 10−(m+1)! ≤ 10−km! = 1/qk m for all m large enough. If we set Uk = � B(p/q, 1/qk) then the set of Liouville numbers if given by �∞ 1 Uk − ratls, and thus: Theorem 6.15 A generic real number is a Liouville number. On the other hand, it is not hard to show that a number chosen at random in [0, 1] is not a Liouville number. 7 Normal spaces There are many separation axioms that interpolate between the Hausdorff spaces and metrizable spaces. In this section we will focus on the most important such condition, normality. 41.

A topological space is said to be normal if (a) points are closed and (b) any pair of disjoint closed sets can be engulfed by a pair of disjoint open sets. More formally, if A and B are closed and A ∩ B = ∅, then there exist open set U and V with A ⊂ U, B ⊂ V and U ∩ V = ∅. Note that a normal space is Hausdorff (take A and B to be single points, which are closed as part of the definition of normality). Any metric space is normal: we can simply let U = and V =. Slightly more subtle is: Theorem 7.1 A compact Hausdorff space is normal. Proof. We first prove X is regular: a closed set A and a point b ̸∈ A have disjoint open neighborhoods. By the Hausdorff assumption, for each a ∈ A we have disjoint neighbor- hoods Ua of a and Va of b. Finitely many of these cover A; call their union U, and let Vb be the intersection of the correspond open sets Vb. Then U and V provide disjoint neighborhood of A and b. We can now prove X is normal by the same style of argument: by regu- larity, given disjoint closed sets A and B, for each a ∈ A we can find disjoint neighborhoods Ua of a and Vb of B. The union U of finitely many of these Ua covers A, and the intersection V of the corresponding open sets Va contains B. Here are two general results we will prove about normal spaces. They are often called Urysohn’s Lemma and Urysohn’s Metrization Theorem. Theorem 7.2 Let A and B be disjoint closed subsets of a normal space X. Then there exists a function f ∈ C(X) such that f|A = 0 and f|B = 1. Theorem 7.3 A normal space with a countable base is metrizable. Proof of Theorem 7.2. Using normality it is easy to inductively construct a sequence of closed sets As indexed by the numbers of the form s = p/2n, 0 ≤ p/2n ≤ 1, such that 1. A0 = A, A1 = X, 42.

2. We have As ⊂ int(At) whenever s < t. 3. We have As ⊂ X − B whenever s < 1. Once this is done, we define f : X → [0, 1] by f(x) = inf. From the conditions above we get f|A = 0, f|B = 1. For continuity, first note that f −1((−∞, a)) = = � s<a As = � s<a int(As) is a union of open sets, hence open. Similarly f −1((b, +∞)) = = � s>b (X − As) is also open. Since these two types of intervals generate the topology on the real numbers, f is continuous. Remark. Even though f|A = 0, we might not have f −1(0) = A, and similarly for B. Maximal ideals. An important consequence of Urysohn’s Lemma is that if K is a compact Hausdorff space, then the functions in C(K) separate the points of K. Consequently any homomorphism of algebras φ : C(K) → R comes from a point evaluation (exercise). Proof of Theorem 7.3. First construct, for every pair of elements in the countable base B, B′ with B ⊂ B′, a function f : X → [0, 1] which is 1 on B and 0 outside B′. Let the countable set of such functions be (f1, f2, . . .), and let φ : X → [0, 1]N be the tautological embedding: φ(x) = (f1(x), f2(x), . . .). We give the target the product topology. Now the Hilbert cube [0, 1]N is certainly metrizable. so to complete the proof it suffices to show that φ is an embedding. 43.

It is clear that φ is continuous. It is also 1 − 1, since if x ̸= y we can find B, B′ as above with x ∈ B and y ̸∈ B′. (Thus if X were compact, we would be done.) We must show that φ : X → φ(X) is an open map. For this, we must show that for any open set U in X and φ(x) ∈ φ(U), there is a neighborhood V of φ(x) in [0, 1]N such that φ(X) ∩ V ⊂ φ(U). (This shows φ(U) is open in the subspace topology.) To do this, choose B and B′ as above with x ∈ B ⊂ B′ ⊂ U, and let fi be the function that is 1 on B and 0 outside B′. Let V ⊂ [0, 1]N be the open set defined by the condition (fi > 1/2). Then if φ(y) ∈ V , we have y ∈ B′ and hence φ(y) ∈ φ(U), which completes the proof. Note. One can also metrize a regular space with a countable base. Compactifications. It is often useful, given a topological space X which is not compact, to try to find a compactification X of X. This means we find a compact space X, and an inclusion X ⊂ X which is a homeomorphism to its image, such that X is dense in X. The problem is similar to that of metric completion, except that the answer is not unique. For example, we can compactify (0, 1) as a circle or as a an interval [0, 1]. Locally compact spaces. Let X be a Hausdorff space. If X has a basis such that B is compact for all B, we say X is locally compact. The one- point compactification X∗ of X is defined by adding the single point ∞ to X, and declaring that any set of the form (X − K) ∪, with K compact, is a neighborhood of infinity. For example, the 1-point compactification of R2 is the 2-sphere. Exercise: show X∗ is compact. Compactification by ends. Another compactification is given by X = X ∪ ϵ(X), where ϵ(X) is the space of ends of X, defined in terms of the components of X−K. Rather than giving a formal definition we just mention a few examples: (i) X ∼= [0, 1] for X = R; (ii) X ∼= Sn ∼= X∗ for X = Rn, n > 1; (iii) X ∼= X ∪ d × (d − 1)N when X is a regular tree of degree d. The Stone-ˇCech compactification. Now suppose X is a normal space. The biggest compactification of X, denoted β(X), is defined as follows. Let 44.

I(X) be the set of all continuous maps f : X → [0, 1]. Then [0, 1]C is compact, by Tychonoff’s theorem. Define ι : X → [0, 1]I(X) by ι(x)f = f(x). Then β(X) is, by definition, the closure of ι(X). The proof that ι is a homeomorphism to its image is similar to the proof of the metrizability theorem above. This compactification is biggest in the following technical sense: any con- tinuous map f : X → K, where K is a compact Hausdorff space, extends to a continuous map β(f) : β(X) → K. The Stone-ˇCech compactification of N. The case X = β(N) is already exotic. Here N has the discrete topology, so every map a : N → [0, 1] is continuous. In other words, C consists of all sequences with 0 ≤ an ≤ 1, and every sequence has a continuous extension to β(N). Let limξ an denote the value of a at a point ξ ∈ β(N) with ξ ̸∈ N. This ‘generalized limit’ is not only linear and defined for all bounded sequences, but it also satisfies lim ξ (anbn) = (lim ξ an)(lim ξ bn). It agrees with the usual limit when an is convergent. In fact the points of β(N) can be identified with the ultrafilters on N. 8 Algebraic topology and homotopy theory The second main topic of this course will be elementary algebraic topology. The idea of algebraic topology is to associate to a continuous topological space X a discrete algebraic object A(X), with the aim of telling spaces apart. For this to work, we want the map X �→ A(X) to be a functor; that is, if f : X → Y is continuous, we should also get a map f∗ : A(X) → A(Y ) that respects the algebraic structure of A(X). Homotopy. Since A(X) is discrete, if we continuously deform f, we will not change f∗. This suggests that A(X) and A(Y ) will be the same if X is homotopy equivalent to Y . Here is a precise definition. 45.

A pair of continuous maps f, g : X → Y are homotopic if there is a continuous map F : X × [0, 1] → Y such that F0 = f and F1 = g. Example. Let S1 = R/Z. There is an intuitive notion of degree for a map f : S1 → S1, with deg(f) = n when f(x) = nx mod 1. It turns out that f and g are homotopic iff they have the same degree. In particular, the identity map is not homotopic to a constant map. One of our goals will be to prove this. Homotopy equivalence. A homotopy equivalence between spaces is a map f : X → Y such that there exists a map g : Y → X with g ◦ f and f ◦ g homotopic to the identity on X and Y respectively. Retraction and contractible spaces. A special case arises when the inclusion A ⊂ X is a deformation retract of X. This means we have a continuous family of maps Ft : X → X such that (i) Ft(x) = x for all x ∈ A, (ii) F0(x) = x, and (iii) F1(x) ∈ A. The final map F1 is called a retraction of X onto A. By continuous, we mean (t, x) �→ Ft(x) is a continuous function on [0, 1]2 For example, the space X = Rn retracts onto the origin A =. In this case we say X is contractible (homotopy equivalent to a point). Example: A theta, a dumbbell and a figure 8 are all homotopy equivalent. The coolest proof of this is to slightly thicken these sets in the plane. All of the thickenings are homeomorphic, and all retract onto the original set. The punctured plane X = R2− retracts onto the unit circle A = S1 ⊂ X. The retraction can be given in polar coordinates by Ft(es, θ) = (e(1−t)s, θ). It is just obtained by retracting each ray through the origin to its intersection with the circle. The punctured torus. It is an interesting an initially unexpected fact that a punctured torus is homotopy equivalent to a bouquet of 2 circles. What can you say for a punctured surface of genus 2? In fact X = S1×S1−B is homeomorphic to the rotary with an underpass shown in Figure 1. If we think of S1 × S1 as a quotient of the square, then it is easy to see that X retracts onto S1 ∨ S1. But X represents a different thickening of this region from the ones coming from a dumbbell or theta or figure eight in the plane. Namely X has only one boundary component, while the other thickenings have three. This reflects the facet that X has a handle — it does not embed in the plane. 46.

9 Categories and paths We begin a more detailed treatment with a quick review of categories and groups. Categories. A category is a collection of objects A, B, C and an association to each pair of objects a family of morphisms (or arrows), Mor(A, B). We are also given a composition law: Mor(A, B) × Mor(B, C) → Mor(A, C). The main axiom is that composition is associative: (f ◦ g) ◦ h = f ◦ (g ◦ h) whenever both sides are defined. We also require that for every object A we have a morphism idA ∈ Mor(A, A) which behaves as the identity. This means f ◦ idA = idB ◦f = f for all f ∈ Mor(A, B). By considering idA ◦ id′ A, we see that idA is unique. Warning: often the collection of all objects of a category is too big to be a set, e.g. it might be the ‘collection’ of all sets. We will ignore this logical difficulty, which can be circumvented by considering a category as a ‘class’. Examples: 1. The category of all sets, with Mor(A, B) as all maps f : A → B. 2. The category of sets, with Mor(A, B) as the set of relations (subsets f ⊂ A × B). 3. The category of all finite–dimensional vector spaces over R, with Mor(A, B) as the set of linear maps. 4. The category of groups, with Mor(A, B) as group homomorphisms. 5. The category of all topological spaces, with Mor(A, B) as the set of continuous maps. 47.

6. The category of groups, with the morphisms as group isomorphisms. 7. A partially ordered set can be made into the objects of a category, where the morphisms all move upwards. 8. The vertices of a directed graph with form a category, with directed paths as morphisms. 9. A group is a category with one object, in which all morphisms are isomorphisms. 10. The matrices Mn(R) form the objects of a category, with Mor(A, B) =. (Any ring with identity could play the role of Mn(R) here.) 11. The cobordism category. In the case of dimension two, the objects are finite unions of circles, and the morphisms are surfaces they bound. Note: Mor(A, B) may be empty, but Mor(A, A) always contains idA. What is a map? When a category is understood, a map generally refers to a morphism in the category. Thus in the sequel, where we work in the category of topological spaces, we will often discuss a map f : X → Y ; the assumption that f is continuous is implicit. Isomorphisms. We say f ∈ Mor(A, B) is an isomorphism if there exists a g ∈ Mor(B, A) such that g ◦ f = idA and f ◦ g = idB. The set of all isomorphisms from A to itself is called the automorphism group Aut(A). By applying associativity to the expression g′ ◦ f ◦ g, we find: The inverse of an isomorphism is unique. We let Isom(A, B) ⊂ Mor(A, B) denote the set of all isomorphisms for A to B, and we let Aut(A) = Isom(A, A). Groups. A group is a set G equipped with a product operation G × G → G denoted f ∗ g, an inverse map G → G, denoted g �→ g−1, and an identity element denoted e ∈ G, such that: 1. e ∗ g = g ∗ e = e for all g ∈ G, 48.

2. g ∗ g−1 = g−1 ∗ g = ea for all g ∈ G, and 3. (f ∗ g) ∗ h = f ∗ (g ∗ h) for all f, g, h ∈ G. In any category, Aut(A) is a group. Conversely, a group is the same thing as a category with one object where all morphisms are isomorphisms. Examples: if A is a set with n elements, then Aut(A) ∼= Sn. If A is a vector space of dimension n over R, then Aut(A) ∼= GLn(R). Proposition 9.1 If A ∼= B then Aut(A) ∼= Aut(B). The second isomorphism in general depends on the choice of a map from A to B. Groupoids. A groupoid is a category in which every morphism is an iso- morphism. Basic examples are: sets with bijections; vector spaces or groups, with isomorphisms; topological spaces with homeomorphisms. The category of paths. (Also known as the fundamental groupoid.) Here is a very useful and perhaps surprising example of a groupoid. Let X be a topological space. Consider the category P(X) whose objects are the points A, B, C . . . ∈ X. A path from A to B is a continuous map f : [0, 1] → X with f(0) = A and f(1) = B. Two paths f0 and f1 are homotopic if there is a continuous family of paths fs(t) from A to B, s ∈ [0, 1], interpolating between them. This notion gives an equivalence relation on paths. We denote the homotopy class of a path by [f], and let Mor(A, B) =. The identity idA = [cA] ∈ Mor(A, A) is the class of the constant path, cA(t) = A. Question. When is Mor(A, B) empty? The interesting part is composition. Once we have a path f from A to B and a path g from B to C, we can form a new path by (f ∗ g)(t) = � f(2t) for t ∈ [0, 1/2], and g(2t − 1) for t ∈ [1, 2, 1]. Actually the map is given by [f] ∗ [g] = [f ∗ g]. 49.

One must check that this map is well–defined. But deformations of f and g paste together to give deformations of f ∗ g, so it is. Figure 4 explains why [cA ◦ f] = [f] and [(f ∗ g) ∗ h] = [f ∗ (g ∗ h)]. f f A f * g f h g * h Figure 4. Proof that the category of paths really is a category. Inverses. Finally we need to check that every path f(t) has an inverse. But this is just that path defined by g(t) = f(1 − t). Then [g ∗ f] ∈ Mor(A, A) is the same as [cA], since g ∗ f can be deformed to a point, like slurping up a single noodle of spaghetti. The category of unparameterized paths. We could also have identified paths f and g if f(t) = g(h(t)) for some homeomorphism h : [0, 1] → [0, 1] fixing the endpoints. The result would still be a category; everything would go through except the existence of inverses. The result is a continuous version of the category of paths in a graph. The invariant π0(X). As in any category, we can classify the objects up to isomorphism, i.e. introduce the equivalence relation A ∼ B if there exists an isomorphism from A to B. For example, finite-dimensional vector spaces over R are classified up to isomorphism by their dimensions. The set of isomorphism classes in the category of paths is called the space of path components: π0(X). Usually when dealing with the fundamental group, we assume our space is path–connected. The invariant π1(X): the fundamental group. The fundamental group of a space, in terms of the category of paths, is now defined by: π1(X, A) = Mor(A, A) = Aut(A). It is simply the group of all loops based at A. 50.