Proving Urysohn's Lemma and the Application to G_δ Sets

[AKG]

This is a problem I am doing for review, in the section of my book on the Urysohn lemma:

A G_{\delta} set in a space X is the intersection of a countable collection of open sets in X. Let X be normal. Prove that if A is a closed G_{\delta} set in X, then there exists a continuous function f : X \to [0, 1] such that f(x) = 0 if x is in A, and f(x) > 0 otherwise.

The Urysohn lemma states:

X is normal, A and B are disjoint closed subsets of X. Let [a,b] be a closed interval in the real line. Then there exists a continuous map

f : X \to [a, b]

such that f(x) = a for every x in A, and f(x) = b for every x in B.

The proof for this lemma has some open set U₀ containing A such that f(U₀) = {0}, and it seems possible that f might be 0 even for some points outside this U₀. But I need a function that is 0 on A, and only on A. Now the proof of the lemma also has a bunch of sets U_p corresponding to each rational p in [0, 1] such that p > q implies \overline{U_q} \subset U_p. The fact that A is a G_{\delta} set tells me that

A = \bigcap _{n = 1} ^{\infty} A_n

I suppose I could also re-write A as:

A = \bigcap _{n=1} ^{\infty} \left ( \bigcap _{k = 1} ^{n}A_k\right ) = \bigcap _{n=1} ^{\infty} B_n

so that A is an intersection of a descending chain of open sets. The thing is, the A_n and/or the B_n are not, a priori, like the U_q in that

(U_p \subset U_q \wedge U_p \neq U_q) \Rightarrow \overline{U_p} \subset U_q

I'm not sure what to do. I was thinking to first look at A and B_1^C as my first two closed sets, and use the Urysohn lemma to define a continuous function from X to [1/2, 1] choosing U_{1/2}^1 to be B_2. Next, look at A and B_2^C as my closed sets, and use the Urysohn lemma to find a function from X to [1/3, 1/2] choosing U_{1/3}^2 to be B_3. In case it's not clear, I essentially plan on applying the Urysohn lemma infinitely many times, and the k^th time I apply it, the smallest open set containing A that is used in defining the function will is what I'm calling U_{(k+1)^{-1}}^k.

I'll then get a bunch of functions, the k^th one mapping B_k+1 to 1/(k+1) and B_k^C to 1/k. I will just look at how these functions behave on \overline{B_{k} - B_{k+1}} and then use the pasting lemma to argue that it is continuous. Does this seem like a workable approach?

 

[AKG]

There are definitely some problems with the approach so far. I'll try to fix it, and then see if this fix is possible. I also don't think that what I have above is all that clear.

A is a G_{\delta} set, so it can be written as the countable intersection \cap A_n. Define:

B_n = \bigcap _{k=1} ^n A_k

Then:

A = \bigcap _{n = 1} ^{\infty} B_n

(B_n) is a descending sequence of open sets. Pick a subsequence (C_n) such that C₁ = B₁, and C_n+1 is the largest element of the B sequence contained in C_n such that

\overline{C_{n+1}} \subset C_n

(* I don't know yet that this is possible, but we'll see later). Define functions

f_n : X \to \left [\frac{1}{n+1},\frac{1}{n}\right ]

that are continuous, and such that f(C_n+1) = {(n+1)^-1} and f(X - C_n) = {n^-1}. Do this by applying the Urysohn lemma to disjoint closed sets A and X - C_n, and picking C_n+1 to be the open set containing A whose closure is contained in X - (X - C_n) (just like the Urysohn lemma, when applied to the closed disjoint sets A, B and the interval [0,1] uses the normality of X to choose an open set U₀ containing A whose closure is contained in X - B).

Consider the functions:

g_1 = f_1|_{X - C_2},\ g_2 = f_2|_{\overline{C_2} - C_3},\ \dots ,\ g_n = f_n|_{\overline{C_n} - C_{n+1}},\ \dots

These functions are continuous on closed sets whose union is X. Two of these functions have overlapping domains when their indices are adjacent. In particular, g_n and g_n+1 have domains that overlap precisely on the boundary of C_n+1 which is

\overline{C_{n+1}} - C_{n+1}

This functions will agree on the overlapping portions, so although the pasting lemma is only for two functions defined on closed sets whose union is X, I would like to say that I can apply to this (possibly infinite) countable set of functions, and thereby get a continuous function from all of X to [0, 1] by pasting all these functions together, to get a function f.

Now f(A) would defined to be {0}, and this would be continuous. Any point outside A would be outside some C_n, and would thus be greater than 0.

Going back to *, suppose such a choice is not possible. Then there is some C_n = B_m such that for all l > m, the closure of B_l is not contained in B_m. But by normality of X, there is some open subset of X, B, such that B contains A, and B_m contains B's closure. So maybe instead I can choose C_{n+1} = B_{m+1} \cap B and replace the sequence

\langle B_l \rangle _{l > m+1}

with

\langle B_l \cap B \rangle _{l > m+1}

and repeat this argument to find the sequence <C_n>.

 

[Hurkyl]

This is a problem I am doing for review, in the section of my book on the Urysohn lemma:
A G_{\delta} set in a space X is the intersection of a countable collection of open sets in X. Let X be normal. Prove that if A is a G_{\delta} set in X, then there exists a continuous function f : X \to [0, 1] such that f(x) = 0 if x is in A, and f(x) > 0 otherwise.

You sure you have the problem right?

For any continuous function f, the zero set of f must be a closed set. (Because it is the inverse image of the closed set {0}) G_{\delta} contains sets that are not closed, so this can't be true.

 

[AKG]

I forgot to write that A is a closed G_{\delta} set.