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Lecture 15: October 20 The Tietze extension theorem. Another important application of Urysohn’s lemma is the following extension theorem for continuous real-valued functions. This result is very useful in analysis. Theorem 15.1 (Tietze extension theorem). Let X be a normal topological space, and A X a closed subset. ⊆ (a) Let I R be a closed interval. Any continuous function f : A I can be extended⊆ to a continuous function g : X I. → → (b) Similarly, any continuous function f : A R can be extended to a contin- → uous function g : X R. → Saying that g extends f means that we have g(a)=f(a) for every point a A. The assumption that A be closed is very important: for example, the function∈ f :(0, ) R with f(x)=1/x cannot be extended continuously to all of R. The∞ proof→ of Theorem 15.1 goes as follows. Using Urysohn’s lemma, we shall construct a sequence of continuous functions sn : X I that approximates f more and more closely as n gets large. The desired function→ g will be the limit of this sequence. Since we want g to be continuous, we first have to understand under what conditions the limit of a sequence of continuous functions is again continuous.

Uniform convergence. Consider a sequence of functions fn : X R from a → topological space X to R (or, more generally, to a metric space). We say that the sequence converges (pointwise) to a function f : X R if, for every x → ∈ X, the sequence of real numbers fn(x) converges to the real number f(x). More precisely, this means that for every x X and every ε>0, there exists N with ∈ fn(x) f(x) <εfor all n N. Of course, N is allowed to depend on x; we get a more| restrictive− | notion of convergence≥ if we assume that the same N works for all x X at the same time. ∈ Definition 15.2. A sequence of functions fn : X R converges uniformly to a → function f : X R if, for every ε>0, there is some N such that f(x) fn(x) <ε for all n N and→ all x X. | − | ≥ ∈ The usefulness of uniform convergence is that it preserves continuity. Lemma 15.3. The limit of a uniformly convergent sequence of continuous func- tions is continuous.

Proof. Suppose that the sequence of continuous functions fn : X R converges → uniformly to a function f : X R. We have to show that f is continuous. Let V → 1 ⊆ R be an arbitrary open set; to show that f − (V ) is open, it suffices to produce for 1 every point x0 f − (V ) a neighborhood U with f(U) V . This is straightforward. One, f(x ) V∈, and so there is some r>0with ⊆ 0 ∈ B f(x ) V. r 0 ⊆ Two, the sequence converges uniformly,￿ and￿ so we can find an index n such that f (x) f(x)

Now we can show that f(U) V . Let x U be any point; then ⊆ ∈ r r r f(x) f(x ) f(x) f (x) + f (x) f (x ) + f (x ) f(x ) < + + = r, | − 0 |≤| − n | | n − n 0 | | n 0 − 0 | 3 3 3 and so f(x) B f(x ) V . ∈ r 0 ⊆ ￿ Proof of Tietze’s￿ theorem.￿ We will do the proof of Theorem 15.1 in three steps. Throughout, X denotes a normal topological space, and A X a closed subset. The first step is to solve the following simpler problem.⊆ Given a continuous function f : A [ r, r], we are going to construct a continuous function h: X R that is somewhat→ − close to f on the set A, without ever getting unreasonably large.→ More precisely, we want the following two conditions: 1 (15.4) g(x) r for every x X | |≤3 ∈ 2 (15.5) f(a) g(a) r for every a A | − |≤3 ∈ To do this, we divide [ r, r] into three subintervals of length 2 r, namely − 3 1 1 1 1 I = r, r ,I= r, r ,I= r, r . 1 − 3 2 −3 3 3 3 1 1 Now consider the two sets B = f − (I1) and C = f − (I3). They are closed subsets of A (because f is continuous), and therefore of X (because A is closed); they are also clearly disjoint.ï Becauseò X is normal,ï Urysohn’sò lemmaï ò produces for us a continuous function h: X [ 1 r, 1 r]with → − 3 3 1 r for x B, f(x)= − 3 ∈ 1 r for x C. 3 ∈ 1 Since f(x) 3 r, it is clear that (15.4) holds. To show that (15.5) is also satisfied, take any| point|≤ a A. There are three cases. If a B,thenf(a) and h(a) both belong to I ;ifa ∈ C,thenf(a) and® h(a) both belong∈ to I ;ifa B C,then 1 ∈ 3 ￿∈ ∪ f(a) and h(a) both belong to I2. In each case, the distance between f(a) and h(a) 2 can be at most 3 r,whichproves(15.5). The second step is to use the construction from above to prove assertion (a) in Theorem 15.1.IfI consists of a single point, the result is clear. On the other hand, any closed interval of positive length is homeomorphic to [ 1, 1]; without loss of generality, we may therefore assume that we are dealing with− a continuous function f : A [ 1, 1]. As I said above, our strategy is to build a uniformly convergent → − sequence of continuous functions sn : X [ 1, 1] that approximates f more and more closely on A. → − To begin with, we can apply the construction in the first step to the function f : A [ 1, 1]; the result is a continuous function h1 : X R with → − → 1 2 h (x) r and f(a) h (a) r. | 1 |≤3 | − 1 |≤3 Now consider the difference f h1, which is a continuous function from A into the 2 2 − closed interval [ 3 r, 3 r]. By applying the construction from the first step again 2 − (with r = ), we obtain a second continuous function h2 : X R with 3 → 1 2 2 2 h (x) and f(a) h (a) h (a) . | 2 |≤3 · 3 | − 1 − 2 |≤ 3

Å ã 3

Notice how h1 + h2 is a better approximation for f than the initial function h1.We can obviously continue this process indefinitely. After n steps, we have n continuous functions h1,...,hn : X R with → 2 n f(a) h (a) h (a) . | − 1 −···− n |≤ 3

By applying the construction to the function f h1 hn and the value n − −···− r =(2/3) , we obtain a new continuous function hn+1 : X R with → 1 2 n Å ã 2 n+1 h (x) and f(a) h (a) h (a) h (a) . | n+1 |≤3 · 3 | − 1 −···− n − n+1 |≤ 3 Now I claim that the function ∞ g(x)= h (x) Å ã n Å ã n=1 ￿ is the desired continuous extension of f. To prove this claim, we have to show that the series converges for every x X; that the limit function g : X [ 1, 1] is continuous; and that g(a)=f(a)∈ for every a A. → − To prove the convergence, let us denote by∈s (x)=h (x)+ + h (x)then-th n 1 ··· n partial sum of the series; clearly, sn : X R is continuous. If m>n,then → m m k 1 n 1 2 − 2 s (x) s (x) h (x) . | m − n |≤ | k |≤3 3 ≤ 3 k=￿n+1 k=￿n+1 This proves that the sequence of real numbers sn(x) is Cauchy; if we define g(x) as the limit, we obtain a function g : X R. Now we can let m go to infinity in the inequality above to obtain → Å ã Å ã 2 n g(x) s (x) | − n |≤ 3

for every x X. This means that the sequence sn converges uniformly to g, and so by Lemma∈ 15.3, g is still continuous. It is also not hard to see that g takes values in [ 1, 1]: for every x X,wehave − ∈ Å ã n 1 ∞ 1 ∞ 2 − g(x) h (x) =1, | |≤ | n |≤3 3 n=1 n=1 ￿ ￿ by evaluating the geometric series. It remains to show that g(a)=f(a)whenever a A. By construction, we have ∈ Å ã 2 n f(a) s (a) = f(a) h (a) h (a) ; | − n | | − 1 −···− n |≤ 3 letting n , it follows that f(a) g(a) = 0, which is what we wanted to show. The third→∞ step is to prove assertion| − (b)| in Theorem 15.1, where we are given a continous function f : A R.Evidently,R is homeomorphicÅ toã the open interval ( 1, 1); the result of the second→ step therefore allows us to extend f to a continuous function− g : X [ 1, 1]. The remaining problem is how we can make sure that g(X) ( 1, 1).→ Here− we use the following trick. Given g, we consider the subset ⊆ − 1 D = g− 1, 1 X. {− }⊆ Because g is continous, this set is closed; it is also disjoint from the closed set A, because g(A) ( , 1). By Urysohn’s lemma, there is a continous function ⊆ − 4

ϕ: X [0, 1] with ϕ(D)= 0 and ϕ(A)= 1 . Now consider the continuous function→ ϕ g. It is still an extension{ } of f, because{ } we have · ϕ(a) g(a)=g(a)=f(a) · for a A. The advantage is that ϕ g maps X into the open interval ( 1, 1): if x D∈,thenϕ(x) g(x) = 0, while if·x D,then ϕ(x) g(x) g(x) <−1. This completes∈ the proof· of Tietze’s extension￿∈ theorem. | · |≤| |

Invariance of domain. The next topic I wish to discuss is a famous result called the invariance of domain; roughly speaking, it says that Rm and Rn are not homeo- morphic unless m = n. This result is of great importance in the theory of manifolds, because it means that the dimension of a topological manifold is well-defined. Re- call that an m-dimensional manifold is a (second countable, Hausdorff) topological space in which every point has a neighborhood homeomorphic to an open subset of Rm. If an open set in Rm could be homeomorphic to an open set in Rn, it would not make sense to speak of the dimension of a manifold. The first person to show that this cannot happen was the Dutch mathematician Brouwer (who later became one of the founders of “intuitionist mathematics”); in fact, he proved the following stronger theorem.

Theorem 15.6 (Invariance of Domain). Let U Rn be an open subset. If f : U ⊆ → Rn is injective and continuous, then f(U) is also an open subset of Rn. Recall that the word “domain” is used in analysis to refer to open subsets of Rn. Brouwer’s result tells us that if we take an open subset in Rn and embed it into Rn in a possibly different way, the image will again be an open subset.

Corollary 15.7. If U Rn is a nonempty open subset, then U is not homeomor- ⊆ phic to a subset of Rm for mtopology; but my plan is to present an elementary proof that only requires the theorems and definitions that we have talked about so far. The broad outline is the following: (1) We prove a combinatorial result about triangulations of simplices, called Sperner’s lemma. 5

(2) From Sperner’s lemma, we deduce Brouwer’s fixed point theorem: every continuous function from the closed ball in Rn to itself has a fixed point. (3) The fixed point theorem can then be used to prove the invariance of domain theorem.