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Finite Spaces Handouts 1 Finite Spaces Handouts 1 Henry Y. Chan/Zhouli Xu June 19, 2012 1 General Topology 1.1 Basic Definitions Definition 1.1.1. A topological space (X; U) consists of a set X and a collection U (called a topology of X) of subsets of X satisfying the following properties. •;;X 2 U [ • fUigi2I ⊆ U ) Ui 2 U (U is closed under arbitrary unions) i2I n n [ • fUigi=1 ⊆ U ) Ui 2 U (U is closed under finite intersections) i=1 Sets in U are called open sets in X. Sets whose complements are in U are called closed sets in X. If x 2 U 2 U, we call U a neighborhood of x. Exercise 1.1.2. Recall the definition of open sets in a metric space (X; d). A subset U ⊆ X is open if for every x 2 U, there exists r > 0 such that Br (x) ⊆ U, where Br (x) = fy 2 X j d (x; y) < rg is an open ball. Show that open sets defined this way satisfy the axioms above, i.e., the two notions of open sets coincide. Fun Exercise 1.1.3. Why aren't open sets closed under arbitrary intersections? Try to find a simple counterex- ample? (Hint: Consider R). Example 1.1.4. For any set X, the discrete topology Udis and the trivial topology Utriv are defined as X Udis = 2 (every subset of X is open) Utriv = f;;Xg In other words, the discrete topology and the trivial topology are the minimal and the maximal topology of X satisfying the axioms, respectively. We will see why these two examples are not "interesting" in Fun Exercise 1.3.3.. Definition 1.1.5. For any infinite set X, the cofinite topology is defined as Ucof = fU j XnU is finite g [ f;g In other words, close sets are either finite sets or X. Exercise 1.1.6. Check that cofinite topology satisfies all the axioms. Moreover, show that no metric on X defines the cofinite topology. This gives an example of a topological space which is not a metric space. 1 1.2 Basis Definition 1.2.1. A basis of a set X is a collection B of subsets of X such that (1) For every x 2 X, there exists B 2 B such that x 2 B. (2) If x 2 B1 \ B2, where B1;B2 2 B, then there exists B3 2 B such that x 2 B3 ⊆ B1 \ B2. We say a topology U is generated by B if for every x 2 U 2 U, there exists B 2 B such that x 2 B ⊆ U. S Proposition 1.2.2. The topology generated by a basis B is just i2I Bi j Bi 2 B . Example 1.2.3. The set of all open intervals f(x; y)gx<y is a basis generating the standard (metric) topology on R. Fun Exercise 1.2.4. Is the set of all open intervals with rational endpoints f(x; y)g also a basis generating x<y;x;y2Q the standard topology on R? If so, do you think this is a better choice of a basis for R? Why? In most situations, it suffices to check facts on a basis rather than on a topology. The following proposition gives a great example. Proposition 1.2.5. On a set X, let U and U 0 be the topologies generated by the bases B and B0, respectively. Then U ⊆ U 0 if and only if for every x 2 B 2 B, there exists B0 2 B0 such that x 2 B0 ⊆ B. 2 Exercise 1.2.6. In R , one might consider the set of open disks (balls) fBr (x; y)g as a basis, while others might prefer the set of open rectangles (boxes) f(x1; y1) × (x2; y2)g. Show that these two bases generate the same 2 topology on R . (Hint: Show that every open ball contains an open box and vice versa.) 1.3 Continuous Maps We have learn the epsilon-delta definition for continuous maps. However, in Exercise 1.1.6., we know that not all topological spaces can be equipped with a metric. Hence we need a more general definition for continuous maps between general topological spaces. Definition 1.3.1. A map f :(X; U) ! (Y; V) is continuous if for every V 2 V, its preimage f −1 (V ) 2 U. Proposition 1.3.2. If X; Y are metric spaces, then the definition for continuous maps above is equivalent to the epsilon-delta definition for continuous maps. Fun Exercise 1.3.3. Recall the definition of the discrete topology Udis and the trivial topology Utriv. Let (Y; V) be an arbitrary topological space. What do you observe from an arbitrary map f :(X; Udis) ! (Y; V)? How about an arbitrary map f :(Y; Y) ! (X; Utriv)? A continuous map f : X ! Y is called a homeomorphism if it is a bijection and its inverse f −1 : Y ! X is also continuous. In this case, we say X and Y are homeomorphic. Fun Exercise 1.3.4. Classify all upper case alphabets up to homeomorpshim. (This is just for fun, really.) ABCDEFGHIJKLMNOPQRSTUVWXYZ 2 1.4 Four Kinds of Induced Topology Definition 1.4.1. (Subspace Topology) Let (X; U) be a topological spaces and A ⊆ X be a subset. The subspace topology on A relative to X is defined as UA = fU \ A j U 2 Ug In other words, the subspace topology is the smallest topology so that the inclusion map i : A ! X is continuous. Definition 1.4.2. (Quotient Topology) Let (X; U) be a topological spaces and f : X ! Y be a surjective map. The quotient topology on Y is defined as −1 UY = O j f (O) 2 U In other words, the quotient topology is the smallest topology so that f is continuous. Exercise 1.4.3. Consider the map f : [0; 1] ! fa; bg defined as a if x = 0 f (x) = b if x 6= 0 Clearly f is surjective. What is the quotient topology on fa; bg? Assume we have an equivalence relation ∼ on a topological space (X; U). Let X= ∼ be the set of equivalence classes. There is a natural map q : X ! X= ∼, which sends each element to its equivalence class. Clearly q is surjective, so we can talk about the quotient topology on X= ∼. Fun Exercise 1.4.4. Define ∼ on R as x ∼ y , x − y 2 Z. The quotient R= ∼ is also denoted as R=Z. Does R=Z look like any topological space that we are familiar with? Definition 1.4.5. (Product Topology) Let (X; U) and (Y; V) be two topological spaces. The product topology on X × Y is generated by the basis Bprod = U × V In other words, the product topology is the smallest topology so that the projection maps pX : X × Y ! X; pY : X × Y ! Y are continuous. Exercise 1.4.6. Let X; Y be topological spaces and A ⊆ X; B ⊆ Y . There are two ways to define a topology on A × B. We can either first take the subspace topologies on A ⊆ X and B ⊆ X and then take their product topology, or first take the product topology on X × Y and then take the subspace topology on A × B ⊆ X × Y . Show that these two ways define the same topology on A × B. Definition 1.4.7. (Disjoint Union (also called Coproduct) Topology) Let (X; U) and (Y; V) be two disjoint topological spaces. The coproduct topology on X q Y is defined as Tq = fU [ V j U 2 U;V 2 Vg In other words, the coproduct topology is the smallest topology so that the inclusion maps iX : X ! X q Y; iY : Y ! X q Y are continuous. Fun Exercise 1.4.8. The assumption X; Y being disjoint is totally unnecessary. For example, we may want to take a disjoint union of two copies of X. Can you come up with a definition that does not require X and Y to be disjoint? (Hint: A disjoint union of two copies of R can be thought as a union of two lines fy = 0g and fy = 1g 2 in R .) 3 1.5 Separation Axioms and Alexandroff Spaces Definition 1.5.1. A topological space (X; U) is T0 if for every x 6= y in X, there exists U 2 U such that U contains one of x; y but not the other. Fun Exercise 1.5.2. Show that X is T0 if and only if for every x 6= y in X, there exists B 2 B such that B contains one of x; y but not the other. In other words, if sets in B cannot distinguish x; y, then neither can sets in U. (Hint: Every set in U is a union of sets in B.) Definition 1.5.3. A topological space (X; U) is T1 if fxg is a closed set for all x 2 X. Equivalently, for every x 6= y in X, there exist U; V 2 U such that x 2 U; y 2 V but x2 = V; y2 = U. (Check that these two definitions are equivalent.) Fun Exercise 1.5.4. Show that the only topology which is both Alexandroff and T1 is the discrete topology. Definition 1.5.5. A topological space (X; U) is T2, or Hausdorff, if for every x 6= y in X, there exist U; V 2 U such that x 2 U; y 2 V and U \ V = ;.
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