Perfectly normal space problem

[radou]

Homework Statement

(1) Every metrizable space is perfectly normal.

(2) A perfectly normal space is completely normal

The Attempt at a Solution

A space X is perfectly normal if it's normal and if every closed set in X is a Gδ set.

(1)

I found an alternative proof, but I find this one shorter and more elegant, so I'd like to check it.

Let (X, d) me a metric space. Let d' be the standard bounder metric defined with d'(x, y) = min{d(x, y), 1}. We know that d' induces the same topology as d. Now, let A be a closed subset of X. Define d'(x, A) the usual way. This function is a continuous function from X to [0, 1] which vanishes precisely on A. Hence A is a (closed) Gδ set in X.

(2)

There is a hint in the book, but I ignored it, and tried to prove it this way (I know there's probably something wrong with this, so I'd like to verify if it's what I think it is):

Let X be perfectly normal. We wish to prove it is completely normal, so let Y be a subset of X, and let A be a closed subset of Y. Then it equals an intersection of a closed subset of X with Y, i.e. A = A'$$\cap$$Y. Let U be a neighborhood of A in Y. Now, here's what I'm not sure about: Since U is open in Y, it must equal the intersection of some open set U' in X with Y (by general properties of unions and intersections, I think this should work). But the problem is that U' doesn't need to be a neighborhood of A' in X, right? Since then we could simply use normality to find an open neighborhood of U' containing A whose closure is contained in U' in X, and hence in Y, too. We wouldn't even need to use the fact that X is perfectly normal. Am I right here?

 

[micromass]

Let (X, d) me a metric space. Let d' be the standard bounder metric defined with d'(x, y) = min{d(x, y), 1}. We know that d' induces the same topology as d. Now, let A be a closed subset of X. Define d'(x, A) the usual way. This function is a continuous function from X to [0, 1] which vanishes precisely on A. Hence A is a (closed) Gδ set in X.

This seems to be correct.

As for the second proof, I found it hard to follow. You probably want to show that the subset Y is normal. Can you tell me in what way you are trying to show that Y is normal? Is it correct that you take a closed set A and a neighbourhood U, and you try to find a neighbourhood V of A such that the closure of V is contained in U. Is that what you're trying to do?

Let X be perfectly normal. We wish to prove it is completely normal, so let Y be a subset of X, and let A be a closed subset of Y. Then it equals an intersection of a closed subset of X with Y, i.e. A = A'$$\cap$$Y. Let U be a neighborhood of A in Y. Now, here's what I'm not sure about: Since U is open in Y, it must equal the intersection of some open set U' in X with Y (by general properties of unions and intersections, I think this should work).

Yes, I think that this works.

But the problem is that U' doesn't need to be a neighborhood of A' in X, right?

This is indeed a problem.

Since then we could simply use normality to find an open neighborhood of U' containing A whose closure is contained in U' in X, and hence in Y, too. We wouldn't even need to use the fact that X is perfectly normal. Am I right here?

Yes, in this case we don't need that X is perfectly normal. But your assumption (U' is a neighbourhood of A') is quite a large assumption...

 

[radou]

Thanks, you've verified exactly what I thought was wrong. I'm doing it using the hint in the book right now, shouldn't be much of a problem.