BASIC NOTIONS AND RESULTS IN TOPOLOGY. 1. Metric spaces. Sets with finite diameter are called bounded sets. For x X and r > 0 the set

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1 BASIC NOTIONS AND RESULTS IN TOPOLOGY 1. Metric spaces A metric on a set X is a map d : X X R + with the properties: d(x, y) 0 and d(x, y) = 0 x = y, d(x, y) = d(y, x), d(x, y) d(x, z) + d(z, y), for a x, y, z X. The pair (X, d) is caed a metric space. Given Y X the diameter of Y is diam(y ) = sup d(x, y). x,y Y Sets with finite diameter are caed bounded sets. For x X and r > 0 the set B(x, r) = {y X : d(y, x) < r} is caed the open ba centered at x and of radius r and B(x, r) = {y X : d(y, x) r} is caed the cosed ba centered at x and of radius r. The set Y X is caed open if for every x Y there exists r > 0 such that B(x, r) Y. Note that arbitrary unions of open sets are (obviousy) open. The set Y X is caed cosed if X \Y is open. Arbitrary intersections of cosed sets are cosed. The cosure of a set Y X is Y = {Z Y : Z is a cosed set }. The set Y X is caed dense in X if Y = X. If X contains a countabe dense subset we say that X is separabe. The set Y X is caed nowhere dense in X if Y contains no open ba. The space X is of the first (Baire) category if it can be written as a countabe union of nowhere dense subsets. The space X is of the second (Baire) category if it isn t of the first category. We say that a sequence (x n ) in X converges to x X if im n d(x n, x) = 0. 1

2 2 BASIC NOTIONS AND RESULTS IN TOPOLOGY If Y X then its cosure Y consists of Y together with the imits of convergent sequences in Y. A sequence (x n ) in X is a Cauchy sequence if im d(x n, x m ) = 0. m,n Convergent sequences are Cauchy sequences but the converse fais in genera. The metric space (X, d) is caed compete if every Cauchy sequence is convergent. Theorem 1.1. (X, d) is compete if and ony if for any decreasing sequence (Y n ) of subsets (i.e. Y n Y n+1 ) which are nonvoid, cosed and satisfy im n diam(y n ) = 0, their intersection consists of exacty one point. Proof. Assume the property in the statement hods true and et (x n ) be a Cauchy sequence in X. Set Y n = {x m : m n}, and note that these sets are cosed, Y n+1 Y n, and diam(y n ) 0. If n 1 Y n = {x} then d(x n, x) diam(y n ) 0, i.e. (x n ) converges to x. Conversey, assume that (X, d) is compete and et (Y n ) be a sequence as in the statement. Pic x n Y n and note that if m n then d(x n, x m ) diam(y n ), which impies that (x n ) is Cauchy. Then (x n ) is convergent and its imit must beong to each Y n since these are cosed sets. If x is another common point of a Y n then from d(x, x ) diam(y n ) n we see that x = x. Every metric space X can be identified with a dense subset of a compete metric space. The atter can be chosen to be unique up to isometries and is usuay caed the competion of X. Theorem 1.2. (Baire) A compete metric space is of the second category. Proof. Assume the contrary, that is, X is compete but X = n=1y n, where Y n is nowhere dense for a n. Tae an arbitrary cosed ba B(x 0, r 0 ) with x 0 X, r 0 > 0. Since this ba cannot be contained

3 BASIC NOTIONS AND RESULTS IN TOPOLOGY 3 in Y 1, there exists B(x 1, r 1 ) B(x 0, r 0 ) with r 1 < r 0 /2 such that B(x 1, r 1 ) Y 1 =. Repeat this procedure inductivey, that is, find B(x 2, r 2 ) B(x 1, r 1 ) with r 2 < r 1 /2 such that B(x 2, r 2 ) (Y 1 Y 2 ) =, and so on. We obtain a decreasing sequence of cosed bas B(x n, r n ) such that B(x n, r n ) ( n j=1y j = and r n < r 0 /2 n. By Theorem 1.1 these bas must have a common point which, by construction cannot beong to any set Y n. Contradiction! A subset Y of the metric space (X, d) is caed precompact or reativey compact if every sequence in Y has a convergent subsequence. A cosed precompact set is caed compact. Compact sets are bounded and cosed as in R n, but the converse is far from being true in this generaity! Exercise 1. If X is compact then it is compete and separabe. Theorem 1.3. A set Y X is compact if and ony if from every open cover {U α } α A of Y we can extract a finite cover. Proof. The same as in the cacuus course. Let (X, d), (Z, ρ) be metric spaces. A function f : X Z is caed continuous at x 0 X if im n f(x n ) = f(x 0 ) whenever (x n ) is a sequence in X which converges to x 0. The function f is continuous on X if it is continuous at each point of X. Note that a function f : X Z is continuous on X if and ony if the preimage of any open subset of Z is open in X. Theorem 1.4. Let (X, d) be a compact metric space. Then every continuous function f : X R is bounded and attains it minimum and maximum in X. Proof. The same as in the cacuus course. Theorem 1.5. (uniform boundedness) Let (X, d), (Z, ρ) be metric spaces and et F be a famiy of continuous functions from X into Z such that for every x X the set {f(x) Z : f F} is bounded in Z. If X is compete then there exists a nonvoid open set U X such that f F f(u)

4 4 BASIC NOTIONS AND RESULTS IN TOPOLOGY is bounded. Proof. Fix z Z and et X n = {x X : ρ(f(x), z) n f F}. Ceary, each X n is cosed and by assumption we have X n = X. n=1 Then by Baire s Theorem ( Theorem 1.2) at east one of these sets wi contain an open ba. Let (X, d) be a compact metric space and et C(X) denote the set of a compex-vaued continuous functions on X. Given f, g C(X) set ρ(f, g) = sup f(x) g(x). x X Exercise 2. Show that C(X), ρ) is a compete metric space and that a sequence converges in this space if and ony if it converges uniformy on X. A bounded set in (C(X), ρ) is sometimes caed a uniformy bounded famiy of functions. A subset F C(X) is caed equicontinuous at x X if for every ε > 0 there exists δ > 0 such that f(x) f(y) < ε whenever d(x, y) < δ and f F! The set F C(X) is caed equicontinuous if it is equicontinuous at a x X. Theorem 1.6. (Arzea-Ascoi) Let (X, d) be a compact metric space. If F C(X) is bounded and equicontinuous then every sequence in F has a convergent subsequence in C(X) (i.e. F is precompact in C(X)). Proof. Let (f n ) be a sequence in F. Reca that X is separabe and fix a countabe dense subset {x 1,..., x n,...}. Since (f n (x 1 )) is a bounded sequence of compex numbers, it has a convergent subsequence (f n 1 (x 1 )). At its turn (f n 1 (x 2 )) has a convergent subsequence indexed by (n 2 ) and by construction, (f n 2 (x j )) converges in C for j = 1, 2. Inductivey we construct ( smaer and smaer ) subsequences (n m ) such that (f n m (x j )) converges in C for 1 j m. Then the diagona subsequence (f n ) converges pointwise on the set {x 1,..., x n,...}.

5 BASIC NOTIONS AND RESULTS IN TOPOLOGY 5 Let ε > 0. Since F is equicontinuous, for every x X there exists δ x > 0 such that Ceary, d(x, y) < δ x f(x) f(y) < ε, f F. X = x X B(x, δ x ), and since X is compact there exist z 1,..., z n X such that N X = B(z j, δ zj ). j=1 For 1 j N choose x nj B(z j, δ zj ) and note that if x B(z j, δ zj ) then f n (x) f n (x) f n (x) f n (x nj ) + f n (x nj ) f n (x nj ) + f n (x nj ) f n (x) Now the first and the third term on the right hand side are aways ess than ε. By the previous argument we concude that there exists ε such that if, ε and 1 j N, which eads to f n (x nj ) f n (x nj ) < ε, f n (x) f n (x) 3ε for a x X and the resut foows. 2. Topoogica spaces A topoogy on a set X is a coection τ of subsets of X with the properties:, X τ, Finite intersections of eements of τ beong to τ, Arbitrary unions of eements of τ beong to τ. (X, τ) is caed a topoogica space and the subsets of X which beong to τ are caed open sets. A neighborhood of a point x X is an open set containing x. A basis of neighborhoods is a coection τ τ such that every open set is a union of eements of τ. The topoogy τ on X is caed a Hausdorff topoogy if for every x, y X, x Y there exist open sets U x, V y with U V =.

6 6 BASIC NOTIONS AND RESULTS IN TOPOLOGY A point x X is a point of contact of the set Y X if every neighborhood of x intersects Y. A set Y X is cosed if X \ Y is open. The cosure Y of Y is the smaest cosed set containing Y and it equas the union of Y and a its points of contact. If (X, τ), (Z, η) are topoogica spaces, a function f : X Z is continuous if f 1 (U) is open in X whenever U is open in Z. Exampe Let X be a set, et {(X α, τ α )} α A be a famiy of topoogica spaces and et {f α } α A be a famiy of functions with f α : X X α. Then the smaest topoogy τ on X for which each function f α, α A is continuous is given by the basis of neighborhoods V = n i=1 f 1 α i (U i ), where n N, α 1,..., α n A and U i τ αi are arbitrary. A concrete exampe of this type is the product topoogy. Let {(X α, τ α )} α A be a famiy of topoogica spaces and consider their cartesian product which is the set of a functions X = α A X α, f : A α A X α with f(α) X α α A. Such functions are aso caed tupes and denoted by (x α ) α A, where we understand that x α X α for a α A. The product topoogy is the smaest topoogy (on the product space) with the property that a projections P β : α A X α X α, P β ((x α ) α A = x β, are continuous. According to the above observation, a basis of neighborhoods for the product topoogy is given by the sets V = {(x α ) α A : x αi U i ), where n N, α 1,..., α n A and U i τ αi are arbitrary. The topoogica space (X, τ) is compact if every open cover of X contains a finite cover. Equivaenty, (X; τ) is compact if any famiy of cosed subsets of X with the property that any finite subfamiy has a nonvoid intersection, must have itsef a nonvoid intersection.

7 BASIC NOTIONS AND RESULTS IN TOPOLOGY 7 A famiy of arbitrary subsets of X with the above property, that is, any finite subfamiy has a nonvoid intersection is said to have the finite intersection property. Exercise 3. Show that (X, τ) is compact if and ony if every famiy of subsets of X with the finite intersection property, has at east a common point of contact. Theorem 2.1. (Tychonov) The cartesian product of compact topoogica spaces is compact with respect to the product topoogy. Proof. We prove that the condition in Exercise 3 hods. Let M be a famiy of subsets of α A X α with the finite intersection property. Use Zorn s emma to concude that there exists a maxima famiy M 0 which has the finite intersection property and contains M. By maximaity it foows that M 0 has the foowing properties: (a) Any finite intersection of sets in M 0 beongs to M 0, (b) A set that intersects any finite coection of subsets of M 0 must beong to M 0. To see (a) note that if A is a finite intersection of sets in M 0 then M 0 {A} has the finite intersection property and contains M, hence, it must equa M 0. The proof of (b) is identica; if B satisfies that property M 0 {B} contains M and has the finite intersection property, hence, it must equa M 0. Given a set S α A X α and α A, et S α = {x α : (x β ) β A } be the projection of S on the α coordinate. Let M α 0 the famiy of such projections of a sets in M 0. Ceary, M α 0 has the finite intersection property and since X α is compact, it foows that M α 0 must have a common point of contact x α X α. Set x = ( x α ) α A and consider a neighborhood V of x as given in our discussion of the product topoogy, that is, a finite intersection of sets of the form V αi = {(x α ) α A : x αi U i }, i = 1,..., n, where U i is an open subset of X αi for a i. Since each U i intersects every set in M α i 0 it foows that each V αi intersects every set in M 0. Then by (b) V αi M 0 and by (a) we obtain that V M 0. Thus by the finite intersection property V intersects every set in M 0, in particuar every set in M. Since the sets V as above form a basis of neighborhoods we concude that x is a point of contact for every set in M and we are done.