Completely normal space problem

[radou]

Homework Statement

Again, this one seems suspiciously easy, so I'd like to check if I'm missing something.

One needs to show that X is completely normal iff for every pair of separated sets A, B in X (i.e. sets such that both Cl(A) and B, and A and Cl(B) are disjoint), there exist disjoint open sets containing them.

The Attempt at a Solution

<==

Let S be a subspace of X, and take disjoint closed subsets A, B of X. It follows that they are separated, so from the hypothesis it follows that they have disjoint open neighborhoods containing them. Hence S is normal.

==>

Let A and B be two separated subsets of X. Since X\Cl(B) is open, and contains all of A, for any a in A there is a neighborhood U of a disjoint from B. A is contained in the union of these neighborhoods, and this union is open and disjoint from B, and contains A. Do this for the set B, and you get an open set containing B and disjoint from A.

But I smell this is not true, since I didn't use the fact that X is completely normal here.

 

[micromass]

Do this for the set B, and you get an open set containing B and disjoint from A.

Hmm, I don't think it's obvious you can also do this for the set B...

 

[micromass]

Sorry of the above post. Of course you can do this for the set B. But there is a problem. Let U be the open set containing A and disjoint from B. Let V be the open set containing B and disjoint from A. You don't necessariliy have that U and V are disjoint. And I do think you want that...

 

[radou]

Why not? If we look at X\Cl(A), for any element of B we can find a neighborhood dosjoint from Cl(A), not?

 

[radou]

Sorry of the above post. Of course you can do this for the set B. But there is a problem. Let U be the open set containing A and disjoint from B. Let V be the open set containing B and disjoint from A. You don't necessariliy have that U and V are disjoint. And I do think you want that...

Ahhh, of course! I should have realized this immediately! It is even pointed out in the proof of Theorem 32.1 :)

 

[micromass]

Hint: follow the hint

 

[radou]

I think I've got it, but it seems a bit to easy. :)

Since X is completely normal, X\(Cl(A)\capCl(B)) is normal. Cl(A) and Cl(B) are closed in this subspace, since they equal the intersection of closed sets with that subspace. So it follows directly that A and B have disjoint open neighborhoods.

 

[radou]

Actually, to be more precise, Cl(A)\cap(X\(Cl(A)\capCl(B))) and Cl(B)\cap(X\(Cl(A)\capCl(B))) are closed in X\(Cl(A)\capCl(B)).

 

[micromass]

Hmm, you seem to be correct, but there are some things bothering me

Using the normality of X\setminus (Cl(A)\cap Cl(B)) gives you open sets U and V of
Cl(A)\setminus (Cl(A)\cap Cl(B)) and Cl(B)\setminus (Cl(A)\cap Cl(B)). But it is not immediately obvious that they are also open sets of A and B.
You'll need to give an argument why A\subseteq Cl(A)\setminus (Cl(A)\cap Cl(B)). It is here that you use the hypothesis.

 

[radou]

The sets I found are open in our subspace Y = X\(Cl(A)\capCl(B)). Since Y is open in X, our sets must be open in X. Y contains both A and B. So do Cl(A)\capY and Cl(B)\capY. And so do the sets I found. Perhaps I misunderstood you in the last post, but I don't see where I'm wrong.

 

[micromass]

Nono, you're completely correct. But it's common practice for math proofs to state where you use the hypotheses. And I just don't see where you use that A and B are separated sets.
It's not that there's anything wrong with the proof

 

[radou]

Yes, you're right, I should have been a bit more precise. Since A is in X\Cl(B), it is in Y, too. The same holds for B. Here I used the fact that A and B are separated.

It's interesting that I spent about one hour thinking about this problem yesterday, and today I solved it immediately. It turns out to be almost trivial!

 

[micromass]

Hmm, that happens to me a lot to. Sometimes I think for days about problems. And when I finally find it, it seems to be almost trivial. It's not very encouraging

 

[radou]

Haha, now that you say that, it turns out to be quite encouraging for me, since I'm no mathematician, at least not by profession

Thanks for the help, as always.