CHAPTER 1.PRELIMINARY onon the dz,E)=id红， By triangle inequality,it is easy to see that d(,E)is Lipschitz continuous,and thus uniformly We aim to p rove the following tietze extension theo rem in R".It actually holds in more general Theorem 1.10(Tietze extension).Let f:E-R be a continuous function defined on a closed set ECRn with f()C,then there erists a continuous function F:R"R satisfying FE=f and F()≤C. Proof.Set A=r八-C-)B=f-)c=八写c Since A and C are two disjoint closed sets,the function =写利：8 is well-defined.It is easy to see that sm(als号zeg” and 回-n创s29z∈E. Repeat the same process for fg with the bound being2,we get loades and Inductively,we get a sequence of continuous function()defined onR"satisfying The former implies that ()converges uniformly to a continuous function,say G()with the latter implies f(r)-G()=0 for EE. 0 Remark 1.11.The point of Tietze extension theorem is that f is defined on a close set.It is not always poss ible to extend a continuous function defined on an open interval.A simple example is f(x)=sin(3),x∈(0,1 Remark 1.12.A continuous function defined on a closed set needs not to be bounded,however continuous extension still exits
12 CHAPTER 1. PRELIMINARY The limit of a sequence of continuous functions which converges uniformly is continuous. A natural and useful function on the Euclidean space is the distance function. Given E ⊂ R n, let d(x, E) = inf y∈E d(x, y). By triangle inequality, it is easy to see that d(x, E) is Lipschitz continuous, and thus uniformly continuous. We aim to prove the following Tietze extension theorem in R n. It actually holds in more general metric space, we leave the exploration to interested readers. Theorem 1.10 (Tietze extension). Let f : E → R be a continuous function defined on a closed set E ⊂ R n with |f(x)| ≤ C, then there exists a continuous function F : R n → R satisfying F|E = f and |F(x)| ≤ C. Proof. Set A := f −1 ([−C, − C 3 ]) B := f −1 ([− C 3 , C 3 ]) C := f −1 ([C 3 , C]). Since A and C are two disjoint closed sets, the function g1(x) := C 3 d(x, A) − d(x, C) d(x, A) + d(x, C) , is well-defined. It is easy to see that |g1(x)| ≤ C 3 ∀x ∈ R n and |f(x) − g1(x)| ≤ 2C 3 ∀x ∈ E. Repeat the same process for |f − g1| with the bound being 2C 3 , we get |g2(x)| ≤ 2C 9 and |f − g1 − g2| ≤ 4C 9 . Inductively, we get a sequence of continuous function gn(x) defined on R n satisfying |gn(x)| ≤ 2 n−1C 3 n and |f − ( Xn i=1 gi)| ≤ 2 nC 3 n . The former implies that {gn(x)} converges uniformly to a continuous function, say G(x) with |G(x)| ≤ C; the latter implies |f(x) − G(x)| = 0 for x ∈ E. Remark 1.11. The point of Tietze extension theorem is that f is defined on a close set. It is not always possible to extend a continuous function defined on an open interval. A simple example is f(x) = sin( 1 x ), x ∈ (0, 1]. Remark 1.12. A continuous function defined on a closed set needs not to be bounded, however continuous extension still exits
1.5.CONTINUOUS FUNCTIONS AND DISTANCE IN METRIC SPACE 13 1.5.1 Hausdorff distance and Gromov-Hausdorff distance Let X be a subset of a metric space,its e-neighborhood is defined as X-Uex{ldz,g≤eh. The Hausdorff distance between two subsets X,Y is defined as du(X.Y)=inffe>OXCY.YCX. It is a pseudometric on all subsets,because d(X,Y)=0 does not necessarily mean X=Y.When restricting to closed subsets,d(,)becomes a metric.To avoid d(X,Y)=oo,we work further with compact subsets. poduoooouo(aooti. .The Hausdorff distance du(,)defines a metric on D(x) .(D(),dn())is compact if is compact. ·(D(x),da(,》is complete if is complete Proof.To show du(,)defines a metric on D(),we need to show 1.Triangle inequality:dn(X.Y)sdn(X.Z)+dn(Z.Y). 2.dn(X,Y)=0 if and only if x=Y. ume du(X,Z)=r and du(Z,Y)=s,for r r and s1 s,we have ZCY and X C Zr C Y+e Similarly,Z and YCZ,which implies Y cZ C Xn+ here ex that d(r,y)=d(,Y).Hence X gy for r< This contradicts e have Y C X.Thus the conclusion of proof to the reader the clos r metric spaces.The idea is to allow isometric motion in an ambient metric space.i:Y is called y),if dx r.r den(X,Y)inf{dn(i(x),j(Y))), where the inf is taken over all metric spaces Z and isometric embeddingsi:Xandj:YZ. Theorem 1.14.dou defines a metric on the space of compact metrie spaces modulo isometries. We state another convenient description of Gromov-Hausdorff distance.A map f:XY is called an cisometry,if
1.5. CONTINUOUS FUNCTIONS AND DISTANCE IN METRIC SPACE 13 1.5.1 Hausdorff distance and Gromov-Hausdorff distance Let X be a subset of a metric space, its -neighborhood is defined as X = ∪x∈X{y|d(x, y) ≤ }. The Hausdorff distance between two subsets X, Y is defined as dH(X, Y ) = inf{ ≥ 0|X ⊂ Y, Y ⊂ X}. It is a pseudometric on all subsets, because dH(X, Y ) = 0 does not necessarily mean X = Y . When restricting to closed subsets,dH(·, ·) becomes a metric. To avoid dH(X, Y ) = ∞, we work further with compact subsets. Theorem 1.13. Let (X , d) be a metric space. Denote by D(X ) the collection of compact subsets of X . Then we have following • The Hausdorff distance dH(·, ·) defines a metric on D(X ). • (D(X ), dH(·, ·)) is compact if X is compact. • (D(X ), dH(·, ·)) is complete if X is complete. Proof. To show dH(·, ·) defines a metric on D(X ), we need to show 1. Triangle inequality: dH(X, Y ) ≤ dH(X, Z) + dH(Z, Y ). 2. dH(X, Y ) = 0 if and only if X = Y . Proof of 1 Assume dH(X, Z) = r and dH(Z, Y ) = s, for r1 > r and s1 > s, we have Z ⊂ Ys1 and X ⊂ Zr1 , which implies X ⊂ Zr1 ⊂ Yr1+s1 . Similarly, Z ⊂ Xr1 and Y ⊂ Zs1 , which implies Y ⊂ Zs1 ⊂ Xr1+s1 . Together we obtain dH(X, Y ) ≤ r1 + s1, since r1 and s1 are arbitrary, the proof is finished. proof of 2 Suppose there exists x ∈ X but x /∈ Y , then d(x, Y ) = δ > 0. Moreover, since Y is a compact, there exists y ∈ Y such that d(x, y) = d(x, Y ). Hence X * Yr for r < δ. This contradicts to dH(X, Y ) = 0. Thus X ⊂ Y , likewise we have Y ⊂ X. Thus the conclusion follows. We leave the rest of proof to the reader. Hausdorff distance measures the closeness of two subsets of a given metric space. GromovHausdorff distance extends this idea to an intrinsic way of measuring distance between two arbitrary metric spaces. The idea is to allow isometric motion in an ambient metric space. i : X → Y is called an isometric embedding of (X, dX) into (Y, dY ), if dX(p, q) = dY (i(p), i(q)), ∀p, q ∈ X. Given two metric spaces X, Y , the Gromov-Hausdorff distance is defined as dGH(X, Y ) = inf{dH(i(X), j(Y ))}, where the inf is taken over all metric spaces Z and isometric embeddings i : X → Z and j : Y → Z. Theorem 1.14. dGH defines a metric on the space of compact metric spaces modulo isometries. We state another convenient description of Gromov-Hausdorff distance. A map f : X → Y is called an -isometry, if
CHAPTER 1.PRELIMINARY 。dx(,x)-dy(f),frl≤,,x∈X ·f(X)isan-met of Y A subset ZCX is an e-net if Proposition 1.15. ·dca(X,Y)<e→f:X→Ya2e-isometry: ·3f:X→Yame-isomelry→dca(X,Y)<2e rem 1.14.The w don(Y)=0 if and that x is isomet to Y.To this end,we first extract a countable dense subs t 5 of X.This e X is compact,there exists a finite set of X which forms a I-net for mnion of these分nG ble dense subset of =13121 set Y.th there sometry (f(1) take a convergent sub-subsequence.Inductively,we find a subsequence of(still denoted byfor simplicity).which converges at each point of S.Suppose the limit function is f.Hence ldx(s,s-dy(fs),fs川=lim ldx(s,s-dy(n(,fn(s)川=0，s,s∈X, which means f preserves metric on S.Since S is a dense subset of X,f has a unique continuous extension king in the other direction,we get a metric 1.5.2 Invariant of domain 441 are distinct.For example,there does not exist continuous one-to-one correspondence.This is the invariance of domain and relates the notion of topological dimension. Let U C R"be an open set andf:UR"is injective Corollary 1.17.R"is not homeomorphic to Rm,for nm. f:XY between two metric spaces is called a homeomorphism if it is .injective and surjective, ·continuous, .its inverse is also continuous. Proof.Suppo n<m and let f:mR be the homeomorphism.Then by adding m-n zeros, .eF(x）=(fa,0,…,0 be an open set.A contradiction to invariance of domain We can also rephrase the proof to the following fact Theorem 1.18.There does not erist a continuous injection from R"to Rm forn>m The converse direction is Theorem 1.19.There erists a continuous surjection from R"to Rm for n <m
14 CHAPTER 1. PRELIMINARY • |dX(x, x0 ) − dY (f(x), f(x 0 )| ≤ , ∀x, x0 ∈ X; • f(X) is an -net of Y . A subset Z ⊂ X is an -net if Z ⊃ X. Proposition 1.15. • dGH(X, Y ) < ⇒ ∃f : X → Y a 2-isometry; • ∃f : X → Y an -isometry ⇒ dGH(X, Y ) < 2. Proof of Theorem 1.14. The nontrivial part is to show dGH(X, Y ) = 0 if and only if X is isometric to Y . One direction is easy. We just need to show the other direction that dGH(X, Y ) = 0 implies that X is isometric to Y . To this end, we first extract a countable dense subset S of X. This can be done as follows. Since X is compact, there exists a finite set of X which forms a 1 n -net for X. The countable union of these 1 n -net is a countable dense subset of X, denoted by S. Assume S = {s1, s2, · · · }. By Proposition 1.15, there exists 1 n -isometry fn : X → Y . Since {fn(s1)}∞ n=1 is a sequence in a compact set Y , thus we can take a convergent subsequence. Now for s2, we can take a convergent sub-subsequence. Inductively, we find a subsequence of fn (still denoted by fn for simplicity), which converges at each point of S. Suppose the limit function is f. Hence |dX(s, s0 ) − dY (f(s), f(s 0 ))| = limn→∞ |dX(s, s0 ) − dY (fn(s), fn(s 0 ))| = 0, ∀s, s0 ∈ X, which means f preserves metric on S. Since S is a dense subset of X, f has a unique continuous extension ˜f, which also preserves the metric. Working in the other direction, we get a metric preserving map ˜g : Y → X. Thus X is isometric to Y . 1.5.2 Invariant of domain From set theoretical point view, R n and R m have same cardinality. However, the one-to-one correspondence is not easy to write down. When taking more structure into consideration, R n and R m are distinct. For example, there does not exist continuous one-to-one correspondence. This is the invariance of domain and relates the notion of topological dimension. Theorem 1.16 (Invariance of domain). Let U ⊂ R n be an open set and f : U → R n is injective and continuous, then f(U) is also open in R n. Corollary 1.17. R n is not homeomorphic to R m, for n 6= m. f : X → Y between two metric spaces is called a homeomorphism if it is • injective and surjective, • continuous, • its inverse is also continuous. Proof. Suppose n < m and let f : R m → R n be the homeomorphism. Then by adding m − n zeros, i.e F(x) = (f(x), 0, · · · , 0), we get an injective continuous map from R m to R n, whose image fails to be an open set. A contradiction to invariance of domain. We can also rephrase the proof to the following fact Theorem 1.18. There does not exist a continuous injection from R n to R m for n > m. The converse direction is Theorem 1.19. There exists a continuous surjection from R n to R m for n < m
1.5.CONTINUOUS FUNCTIONS AND DISTANCE IN METRIC SPACE 6 The fam Whedn the actrcomto the more aiar facts from ca algebra. Proposition 1.20.There does not erist a linear injection from R"to Rm forn>m. Proposition 1.21.There does not erist a linear surjection from from R"to Rm for n<m. ◇
1.5. CONTINUOUS FUNCTIONS AND DISTANCE IN METRIC SPACE 15 The famous Peano curve provides such an example. When adding the linear structure into account, we come to the more familiar facts from linear algebra. Proposition 1.20. There does not exist a linear injection from R n to R m for n > m. Proposition 1.21. There does not exist a linear surjection from from R n to R m for n < m
CHAPTER 1.PRELIMINARY
16 CHAPTER 1. PRELIMINARY