Topology: Munkres - Urysohn lemma

[Fisicks]

Hi, the problem I am referencing is section 33 problem 4.

Let X be normal. There exists a continuous function f: X -> [0,1] such that fx=0 for x in A and fx >0 for x not in A, if and only if A is a closed G(delta) set in X.

My question is about the <= direction.

So let B be the collection of open sets whose intersection is A and index them with the natural numbers. Let U_n be an element of B. For each U_n, define a function f_n. To define f_n follow the proof of Urysohn lemma using A=A, B=X-U_n.

Define fx= sup{f_n(x)} for all n.

Clearly fx=0 iff x is in A. My problem is with showing continuity. Part of me thinks that if x is an element of X and (a,b) is a basic open set of fx, then there exists an open set U such fU is contained in (a,b) since each f_n is continuous.

 

[fresh_42]

I think we need a reference here as Urysohn doesn't make a statement about the values of $f$ outside $A$ and $B$. Also "G(delta)" needs to be explained.

 

[math771]

However, I am not sure how to show this.



Hi there,

Thank you for your post. I understand your question and will try to provide an explanation.

First, let's define the function f_n as follows:
f_n(x) = d(x, X-U_n) / [d(x, X-U_n) + d(x, A)]

where d(x, X-U_n) represents the distance from x to the complement of U_n in X, and d(x, A) represents the distance from x to the set A.

Now, let's take an arbitrary point x in X and an open interval (a,b) containing fx. We need to show that there exists an open set U containing x such that f(U) is contained in (a,b).

Since f_n is continuous, for each n, there exists an open set V_n containing x such that f_n(V_n) is contained in (a,b). Now, let's define U as the intersection of all V_n's. Since U is an intersection of open sets, it is also open.

For any point y in U, we have d(y, X-U_n) < d(y, X-U_n) + d(y, A) for all n, which means that f_n(y) > 0 for all n. Therefore, sup{f_n(y)} > 0.

Now, let's consider the point x itself. Since x is in U, it must also be in V_n for all n. Therefore, f_n(x) is contained in (a,b) for all n. This means that sup{f_n(x)} is also contained in (a,b).

Hence, we have shown that for any x in X and any open interval (a,b) containing fx, there exists an open set U containing x such that f(U) is contained in (a,b). This proves the continuity of f.

I hope this helps. Let me know if you have any further questions.