HSE/Math in Moscow 2014-2015// Topology 1 // Week 5 lecture notes: metric spaces The Stone-Weierstrass approximation theorem Let X be a topological space. Recall that in Tutorial 3 we have defined the space C b (X) of bounded continuous functions X → R, equipped with the topology induced by the sup norm ||f || = supx∈X |f (x)|. We have shown that C b (X) is complete. If X is compact, then every real-valued continuous function on X is bounded and the sup is always attained at some point, so we can replace sup by max and conclude that the space C(X) of all continuous functions X → R is complete with respect to the max norm. Below, when we write e.g. f ≥ 0 for an f ∈ C(X), this means that f (x) ≥ 0 for all x ∈ X. Theorem 1 (Stone-Weierstrass) Let A ⊂ C(X) be an R-algebra which is 1. closed and 2. distinguishes any two points of X. The latter condition means that for all x, y ∈ X such that x 6= y there is an f ∈ A such that f (x) 6= f (y). Then A = C(X). Application. By taking X = [0, 1] or, more generally, a compact subset of some Rn and A to be the closure of the algebra of all polynomial functions on X, we see that for any ε > 0 and any continuous function f : X → R there exists a polynomial p : X → R such that |f (x) − p(x)| < ε for all x ∈ X. Example. Note that the second condition is important in order for the conclusion to hold: take e.g. X = [0, 1] and A to be the algebra of all continuous functions which are constant on [0, 21 ]. Proof. 1. Take an element f ∈ A. Let us show that then |f | ∈ A. Firstly, note that for all a, b ∈ R such that a ≥ 0, b > 0 we have √ √ √ b b a+b− a= √ √ ≤ √ = b, a+b+ a b so r 1 p 1 f2 + − f2 ≤ √ → 0 n n q p as n → ∞, which means that limn→∞ f 2 + n1 = f 2 = |f |. Since A is assumed closed, to show |f | ∈ A it would √ suffice to show that g ∈ A, provided g : X → R is in A and satisfies 0 < a ≤ g(x) ≤ c for some a, c ∈ R>0 and all x ∈ X. Let us show√this. p √ p √ We have g = c gc = c 1 + (−1 + gc ). Set h = −1 + gc . Since 0 < a ≤ g ≤ c, we have 0 < ac ≤ gc ≤ 1, so −1 < −1 + so using the Taylor expansion for √ a g ≤ −1 + = h ≤ 0, c c 1 + x we get √ 1 1 1 √ c(1 + h − h2 + h3 + · · · ) = g 2 8 16 with the series converging uniformly since h(x) ∈ [−1 + ac , 0] for all x ∈ X . All the terms of the series are in A, hence so is the sum. 2. If f, g ∈ A, then so are max(f, g) = f + g − |f − g| f + g + |f − g| and min(f, g) = . 2 2 3. Suppose Y, Z ⊂ X are two closed subsets. Let us show that there is a h : X → [0, 1] such that h|Y = 1 and h|Z = 0. First, let us fix a y ∈ Y and find a function hy ∈ A such that hy (y) = 1, hy (z) = 0 for all z ∈ Z and hy (x) ≥ 0 for all x ∈ X. We know that, given a z ∈ Z there is an hy,z ∈ A such that hy,z (y) = 1, hy,z (z) = 0. Every z ∈ Z has an open neighbourhood Uz such that hy,z (z 0 ) < 21 forall z 0 ∈ Uz . Choose a finite open subcover {Uz1 , . . . , Uzk } of {Uz }. For every i = 1, . . . , k the function 2(max hy,zi , 21 − 21 ) is identically zero on Uzi and equals 1 at y. So seting k Y 1 1 hy = 2(max(hy,zi , ) − ) 2 2 i=1 1 we get a function hy such that hy |Z = 0, hy (y) = 1. Replacing if necessary hy with max(hy , 0) we can ensure that hy ≥ 0. Now, for all y ∈ Y there is an open neighbourhood Vy 3 y such that hy (y 0 ) > 21 for all y 0 ∈ Vy . Again, choose a Pl finite open subcover {Vy1 , . . . , Uyl } of {Vy }. The function 2 i=1 hyi is zero on Z, is ≥ 0 everywhere in X and is ≥ 1 everywhere in Y , so if we set ! l X h = min 2 hyi , 1 i=1 we get a function ∈ A which is again non-negative everywhere, equals 1 on Y and 0 on Z. Moreover, g ≤ 1. 4. Let us now show that given an arbitrary f ∈ ||f (x) − g(x)|| ≤ 32 ||f ||. Since f is continous and X Set Z = {x ∈ X | f (x) ≤ C(X) such that f ≥ 0 there is a g ∈ A such that f − g ≥ 0 and is compact, there exists a maxx∈X f (x), which we denote as M . 1 2 M }, Y = {x ∈ X | f (x) ≥ M }. 3 3 Both these sets are closed and let h be the function constructed in 3. Set g = M 3 h. For x ∈ Z we have f (x) = f (x − g(x) ∈ [0, 13 M ] and for x ∈ Y we have f (x) − g(x) ∈ [ 31 M, 23 M ]. Finally, for x ∈ X \ (Y ∪ Z) we have f (x) − g(x) ∈ [0, 23 M ] (since 0 ≤ h ≤ 1). So g has all the required properties. 5. Let us now take an arbitrary f ∈ C(X) such that f ≥ 0 again and form a sequence (gn ) of functions in A constructed as follows: g1 is constructed as in 4. starting from f , g2 is constructed again as in 4. but starting from f − g1 instead of f , and generally, gn+1 is constructed as in 4. by replacing f with f − g1 − · · · − gn . We have 2 ||f − g1 − · · · − gn || 3 n n for all n, so ||f − g1 − · · · − gn || ≤ 23 ||f || for all n. As n → ∞, 23 can be made arbitrarily small, so f is in the closure of A, which means it is in A. Finally, an arbitrary f ∈ C(X) can be written as f = (f + C) − C with C being a constant function such that f + C ≥ 0. Since f + C has just been shown to be in A, we see that f ∈ A.♦ ||f − g1 − · · · − gn+1 || ≤ Compactness criteria for metric spaces Theorem 2 For a metric space (X, d) the following statements are equivalent. 1. X is compact. 2. Every sequence of elements of X has a convergent subsequence. 3. X is complete and totally bounded (i.e., for all ε > 0 there is a finite subset K(ε) ⊂ X such that every x ∈ X is < ε away from an element of K(ε); such a set K(ε) is called an ε-net). 4. Every infinite subset of Y ⊂ X has a limit point (i.e., there is an x ∈ X such that any neighbourhood of it contains infinitely many elements of Y ). 5. Every sequence Y1 ⊃ Y2 ⊃ · · · ⊃ Yn ⊃ · · · of non-empty nested closed subsets of X has non-empty intersection. 6. X is countably compact, i.e. every countable open cover of X has a finite subcover. Proof. We prove this by showing that 1 ⇒ 2, 2 ⇒ 3, . . . , 6 ⇒ 1. 1 ⇒ 2 : Let (xn ) be a sequence without a convergent subsequence. As shown in homework 5, question 3b, this means that for every x ∈ X there is an open neighbourhood U (x) 3 x such that U (x) 63 xi for all i sufficiently large. Choose a finite subcover {U (x1 ), . . . , U (xk )} of {U (x)}x∈X . There is an N such that for all n ≥ N the element xi does not belong to any of the sets U (x1 ), . . . , U (xk ), which is a contradiction, as the sets cover X. 2 ⇒ 3 : First let us show that, assuming 2., the space X is complete. Let (xn ) be a Cauchy sequence in X and let us take a convergent subsequence (xnk ) of (xn ). Let x = limk→∞ xnk . Take an ε > 0. We know that there is an N such that for all m, n ≥ N we have d(xm , xn ) < 2ε and for all k ≥ N we have d(xnk , x) < 2ε . So for all m ≥ N we have d(xm , x) ≤ d(xm , xnm ) + d(xnm , x) < 2ε + 2ε = ε. This shows that x = limn→∞ xn . Now let us show that, again assuming 2., X is totally bounded. Take an ε > 0. We want to find a finite ε-net. Let us try to form a sequence (xn ) of elements of X such that d(xi , xj ) ≥ ε for all i 6= j. Let x1 be an arbitrary element of X. Suppose we have already constructed x1 , . . . , xn such that d(xi , xj ) ≥ ε for i 6= j, i, j = 1, . . . , n. Then, if there is an x ∈ X such that d(x, xi ) ≥ ε for all i = 1, . . . , n we set xn+1 = x, and if there isn’t, we stop: {x1 , . . . , xn } is a finite ε-net. So either we will stop at some point and X has a finite ε-net, or there is a sequence (xn ) such that d(xi , xj ) ≥ ε for all i 6= j. In the latter case for i 6= j the elements xi and xj cannot be in U (x, 3ε ) for any x ∈ X, or else d(xi , xj ) ≤ d(xi , x) + d(x, xj ) < 23 ε. So using question 3b of homework 5 again we see that (xn ) does not have a convergent subsequence. 6 ⇒ 1 : For an ε > 0 let A(ε) be a maximal (with respect to inclusion) subset of X such that d(x, y) ≥ ε for all x, y ∈ A(ε) such that x 6= y. Such a subset of X exists by Zorn’s lemma. Note that, because of the 2 S maximality of A(ε), we have x∈A(ε) U (x, ε) = X, so {U (x, ε)}x∈X is an open cover. Note also that if x0 ∈ A(ε) then S x0 6∈ x∈A(ε),x6=x0 U (x, ε), as d(x0 , x) ≥ ε if x ∈ A(ε) and x 6= x0 . So {U (x, ε)}x∈X,x6=x0 is not an open cover. 0 0 S Let us show that A(ε) is finite. If it is not, then it has a countable subset A . Then U (x, ε) for x ∈ A and x∈A(ε),x6∈A0 U (x, ε) together form a countable open cover of X which has no proper subcover (meaning that if one discards any of the open sets of the cover, the remaining subsets no longer cover X), which contradicts the countable compactness of X. S Note that, since any x ∈ X is < ε away from an element of A(ε), the union A = n∈Z>0 A( n1 ) is dense. Note also that it is at most countable. We claim that U (x, q) for x ∈ A and q ∈ Q form a base of the topology of X. (Recall from homework 4 that a family U of open subsets of a topological space X is a called a base iff for any x ∈ X and any open V 3 x there is a U ∈ U such that x ∈ U ⊂ V .) Indeed, take an x ∈ X and an open set U 3 X. We know that there is an r > 0 such that U (x, r) ⊂ U . Let x0 be an element of A such that d(x, x0 ) < 3r . Take a rational number q ∈ ( 3r , 2r ). The open ball U (x0 , q) contains x and for all x00 ∈ U (x0 , q) we have d(x00 , x) ≤ d(x00 , x0 ) + d(x0 , x) < 2r + 3r < r, so U (x0 , q) ⊂ U (x, r). So the sets U (x, q) for x ∈ A and q ∈ Q do form a base. Now, if U is an arbitrary open cover, then set V to be the set of those U (x, q), x ∈ A, q ∈ Q which are included in some U ∈ U. Since U is an open cover and any open subset of X is the union of some number of U (x, q) for x ∈ A and q ∈ Q we see that V is also an open cover. It has a countable subcover, as there are at most countably many U (x, q) for x ∈ A, q ∈ Q. So V has a finite subcover by the countable compactness of X, and so has U, as any V ∈ V is a subset of some U ∈ U.♦ (The remaining part of the proof will be posted later.) 3