Proving Gδ Sets in T1 Spaces | First-Countable T1 Space Problem

[radou]

Homework Statement

Show that in a first-countable T1 space, every one-point set is a Gδ set.

The Attempt at a Solution

This is a problem I was trying to solve a while ago, but didn't manage to. Here's an idea.

Let X be a first countable T1 space, and let x be a point of X. Let y1 be a point different from x (can I choose such a point unless assuming that X has more than one point?). Then X\{y1} is an open neighborhood of x, since {y1} is closed. Since X is forst countable, choose an element B1 contained in X\{y1} containing x. Now, choose an element in B1, say y2, different from x. Since B1\{y2} is open in B1, and B1 is open in X, B1\{y2} is open in X, so one can choose a neighborhood B3 of x contained in B1\{y2}. And so on. Then {x} is contained in the intersection of the Bi's.

Now, assume there is an element y of X, different from x, contained in \cap_{i}Bi. Then y is in Bi, for every i. Now go back to the procedure from the first paragraph, and for B1\{y}, choose a neighborhood B2 of x contained in B1\{y}. But this is a contradiction, since y is in Bi for every integer i. (This is something I'm not really sure about, either - the two choices from different elements from the countable basis for x - can it happen that these two families are completely disjoint? or does the contradiction work, i.e. must there be at least one neighborhood basis elements in common for both procedures?)

This may be a bit confusing, but I'm interested in it works.

 

[radou]

Actually, I think the answer is positive, since this specific procedures produces neighborhoods which are ordered by a propper inclusion, so all of the countable basis elements must come in play. Hm..

 

[micromass]

Hmm, I don't really understand what you did in your second paragraph...

But anyways, what happens if I just pick the y_i very wrong?

Let's say our point is 0.
Let y₁=10 (you picked it arbitrary, so this is a good pick). For B1 I pick ]-2,2[
Let y₂=1.9. So for B2 I pick ]-1-1/2,1+1/2[=]-1.5,1.5[
Ley y₃=1.4. So for B3 I pick ]-1-1/4,1+1/4[=]-1.25,1.25[
Let y₄=1.2. So for B4 I pick ]-1-1/8,1+1/8[

Do you see what I'm trying to do? My sequence of B_n is ]-1-1/2^n,1+1/2^n[. And the intersection is not {0}.

The problem is that I picked the y_i terribly wrong...

 

[radou]

Hm, I see, but what I was trying to say is: the choice of yi's generates one countable family of neighborhood basis elements, ordered by inclusion.

Now if I assume there is some y, which is in the intersection (and different from x), then y is in every basis element generated by the choices of yi's. So, y is in B1. For B1\{y}, choose an element B2 (or B2' - that's the point?) containing x and contained in B1\{y}. In theory, I assumed the existence of this "one and only" countable family which forms a countable basis at x. If this is to, elements must reoccur, contradicting the fact that y is in every one of them.

Perhaps I'm missing something.

In your case, it seems I can't apply what I was trying to do here.

 

[radou]

Yes, I see now that your example shows that this can't work.

It was naive of me to think that this intersection could equal {x}. I'll try to attack this from a different angle.

 

[micromass]

No, your approach is actually a very good one... But instead of picking one y_i after another, maybe you should consider all the points at once...

 

[radou]

By the way, there's something I don't get about your example.

You chose the point 0. But the set Bn is not a countable neighborhood basis for 0. For example, there is no positive integer n such that <-1-1/2^n, 1+1/2^n> is contained in <-1/4, 1/4>.

 

[micromass]

Yeah, but I can extend it to a countable base:

\{]-1-1/2^n,1+1/2^n[~\vert~n\}\cup \{]-1/n,1/n[~\vert~n\}

this is a countable neighbourhood base...

 

[radou]

Hmm, ok. Thinking, thinking...

 

[micromass]

I'll give a hint (spoilered because you may not want it):

Spoiler

Let \mathcal{B}(x) be a countable neighbourhood base.
You need a countable family whose intersection is {x}. \mathcal{B}(x) is a countable family...

 

[radou]

OK, but I'll try not to consider it yet

 

[radou]

By the way, this just occurred to me:

In my "procedure", I chose x in X, and then some y1 different from x in order to obtain a neighborhood X\{y1} of X. I did this to rule out the possibility that X alone is this neighborhood. But what guarantees that the neighborhood from the countable basis at x is not X\{y1} again? Does the countability of this family guarantee it? This may seem like a stupid question, but I'm curious. A family B is countable if there is an injection from B to the set of positive integers. Theoretically, it's possible that this family consists of one element only, right

 

[micromass]

Not sure I understand what the problem is.
Of course it is possible that the neighbourhood is X\setminus \{y_1\} again. But this cannot be the only set in the countable neighbourhood base (unless X has only two elements). Indeed, we can pick X\setminus \{y_1,z\}, this is an open set by the T1-property. Thus there must be a base-set which is contained by it. Thus, this proves that X\setminus\{y_1\} is not the only base-set.

I'm sorry if I misunderstood you...

 

[radou]

Oh, yes...another easy thing I didn't see :) no, you didn't misunderstand

 

[radou]

Btw, if X was countable, your last post gives almost a direct proof, am I right? (just curious)

 

[micromass]

Hmm, good point. It does give a direct proof for countable spaces... I didn't notice that...

 

[radou]

Here's an idea, finally.

Let B be a countable neighborhood basis for x. Then {x}\subseteq \cap_{i}B_{i}. Now let y be a point in X different from x, and assume y\in \cap_{i}B_{i}. Now, let U be a neighborhood of x not containing y (it exists, since X is T1). Then there exists some element of B contained in U, but not containing y, contradicting the fact that y is in the intersection of the Bi's.

And I'll admit - I looked at the spoiler, and that idea occurred to me before. But I just didn't think it through to the end. :)

 

[micromass]

This is correct!

 

[radou]

Again, this was indeed not easy...I over-complicated it. It should have occurred to me intuitively to go for this very intersection :) Thanks.