- #1

AKG

Science Advisor

Homework Helper

- 2,565

- 4

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

The Urysohn lemma states:

The proof for this lemma has some open set U

[tex]A = \bigcap _{n = 1} ^{\infty} A_n[/tex]

I suppose I could also re-write A as:

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

so that A is an intersection of a descending chain of open sets. The thing is, the A

[tex](U_p \subset U_q \wedge U_p \neq U_q) \Rightarrow \overline{U_p} \subset U_q[/tex]

I'm not sure what to do. I was thinking to first look at A and [itex]B_1^C[/itex] as my first two closed sets, and use the Urysohn lemma to define a continuous function from X to [1/2, 1] choosing [itex]U_{1/2}^1[/itex] to be [itex]B_2[/itex]. Next, look at A and [itex]B_2^C[/itex] as my closed sets, and use the Urysohn lemma to find a function from X to [1/3, 1/2] choosing [itex]U_{1/3}^2[/itex] to be [itex]B_3[/itex]. In case it's not clear, I essentially plan on applying the Urysohn lemma infinitely many times, and the k

I'll then get a bunch of functions, the k

*A [itex]G_{\delta}[/itex] 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 [itex]G_{\delta}[/itex] set in X, then there exists a continuous function [itex]f : X \to [0, 1][/itex] 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*

[tex]f : X \to [a, b][/tex]

such that f(x) = a for every x in A, and f(x) = b for every x in B.[tex]f : X \to [a, b][/tex]

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 U

_{0}containing A such that f(U_{0}) = {0}, and it seems possible that f might be 0 even for some points outside this U_{0}. 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 [itex]\overline{U_q} \subset U_p[/itex]. The fact that A is a [itex]G_{\delta}[/itex] set tells me that[tex]A = \bigcap _{n = 1} ^{\infty} A_n[/tex]

I suppose I could also re-write A as:

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

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[tex](U_p \subset U_q \wedge U_p \neq U_q) \Rightarrow \overline{U_p} \subset U_q[/tex]

I'm not sure what to do. I was thinking to first look at A and [itex]B_1^C[/itex] as my first two closed sets, and use the Urysohn lemma to define a continuous function from X to [1/2, 1] choosing [itex]U_{1/2}^1[/itex] to be [itex]B_2[/itex]. Next, look at A and [itex]B_2^C[/itex] as my closed sets, and use the Urysohn lemma to find a function from X to [1/3, 1/2] choosing [itex]U_{1/3}^2[/itex] to be [itex]B_3[/itex]. 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 [itex]U_{(k+1)^{-1}}^k[/itex].I'll then get a bunch of functions, the k

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