Discrete T1 space vs. locally finite basis

[radou]

Homework Statement

The formulation of the problem confused me a little, so just to check.

No T1 space has a locally finite space unless it is discrete.

The Attempt at a Solution

This means that, if X is a discrete T1 space, it has a locally finite basis, right?

Btw, for the converse (which I guess isn't what we need to prove here, but nevertheless) it's almost trivial, and we don't need the T1 condition. Since, if X is a discrete space, then the basis for X is B = {{x} : x is in X}, and hence, if x is in X, the neighborhood {x} of x intersects B in exactly one element, namely {x} itself.

Now, about going for the other direction, I need to show somehow that X must be discrete, if it is T1 and has a locally finite basis.

Let x be in X. There exists a neighborhood Ux of x which intersects the locally finite basis B in finitely many elements, call them B1, ..., Bn. I don't directly see how to conclude that X is discrete, i.e. I'd need to show somehow that the open sets in X are elements o P(X) (I don't think this is likely to be shown), or to show somehow that the basis is exactla the collection B = {{x} : x is in X}, since this is the basis for the discrete topology.

And somehow I need to use the T1 axiom.

This is either very easy and I don't see something obvious, or it's not so easy. Anyway, a small push in the right direction is welcome.

 

[micromass]

Wow, that was not exactly the easiest problem...

Anyway, take \mathcal{B} the locally finite base. Take x an arbitrary point. You'll need to prove that {x} is open.
Now, the trick is to consider \bigcap\{B\in \mathcal{B}~\vert~x\in B\}. So take all the basis sets around x, and consider the intersection of it.
Now, try to show that it's a singleton. This will help you in showing that {x} is open...

 

[radou]

OK, I think I got it. Btw, this is not the first time I came up with a solution after actually going to bed and "calling it a day".

Let x be in X. Let B' be the collection of all basis elements which contain x. Since the basis for X is locally finite, choose a neighborhood of X which intersects finitely many basis elements. Clearly the collection B' must be finite, in the contrary we would have a contradiction. Hence, the intersection S of all elements of B' is open and contains x.

Now, assume there is an element y in S, different from x. Since X is T1, choose a neighborhood U of x disjoint from {y}. Now intersect U with some element from B'. For this intersection, choose a basis element containing x and contained in this intersection. It follows that this basis element belongs to B'. But this is a contradiction, since we assumed y belongs to S, i.e. to every element of the collection B'.

Hence, {x} is open in X, so X has the discrete topology.

 

[micromass]

Yes, that is completely correct!

And you're right, sleep can be the greatest problem solver in mathematics

 

[radou]

Excellent, thanks!