1 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) <r/3 for every x X.Three,f is continuous, and so there is an | n − | ∈ n open set U containing x0 with f (U) B f (x ) . n ⊆ r/3 n 0 2 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 .