Is (R, U) Metrisable? Investigating the Nagata-Smirnov Theorem

[radou]

Homework Statement

Let B = {{x} : x is in R\{pi}}U{R\K : K is a finite subset of R and pi is not in K}U{0} be a family of subsets of R. One needs to prove that B is a basis of some topology U and that (R, U) is not metrisable.

The Attempt at a Solution

Now, I managed to prove that B is a basis, but I'm not really sure about the metrisation issue.

With the material covered in the lectures I'm working with, I couldn't find a way to prove it, so I did some research and found the Nagata-Smirnov theorem which states that a space is metrisable iff it is regular and has a countably locally finite basis.

Now, I assumed that (R, U) is metrisable, hoping to arrive at a contradiction. It is not hard to prove that it is regular, so the only thing left is the countably locally finite basis issue. So, assume B is a union of a countable family of locally finite collections of subsets of R. Let x be some element in R. Then x has a neighbourhood which intersects countably many subsets of B (I'm not really sure about hsi conclusion). But it seems that every neighbourhood of x intersects B in an uncountable number of sets, i.e. whatever the neighbourhood looks like, it always intersects an uncountable number of the sets R\K (I'm not sure about this, either).

I really need a push here, thanka in advance.

 

[Office_Shredder]

Is the 0 there supposed to be the empty set? Otherwise I don't see how it's not covered by the set containing every singleton that's not pi.

We know that R-{pi} has the discrete topology induced from this basis (since every singleton is open), which is metrizable (the distance between every pair of points is 1). So any contradiction is going to occur around the point pi.

I don't think you're going to need the Nagata-Smirnov theorem here; the topology of sets including pi is the cofinite topology, which is kind of terrible. The only open sets containing pi are those whose complement is finite; what does this say about how close almost every point is to pi?

 

[radou]

Is the 0 there supposed to be the empty set? Otherwise I don't see how it's not covered by the set containing every singleton that's not pi.

Yes, 0 denotes the empty set.

We know that R-{pi} has the discrete topology induced from this basis (since every singleton is open), which is metrizable (the distance between every pair of points is 1). So any contradiction is going to occur around the point pi.

I don't think you're going to need the Nagata-Smirnov theorem here; the topology of sets including pi is the cofinite topology, which is kind of terrible. The only open sets containing pi are those whose complement is finite; what does this say about how close almost every point is to pi?

Yes, the only open sets containing pi are of the form R\K, where K is come finite subset of R not containing pi. I don't quite see what you're trying to say.

By the way, an open set from U either has the form of a singleton, or R\K, or R\{pi}, or some subset of R\{pi}, right?

 

[Office_Shredder]

Yes, you've characterized the open sets. Of course R\{pi} is of the form R\K with K finite.

The point is this. If you have a metric space, then in particular the point pi is some distance from every single point.

Let's look a bit more at pi. Let's suppose we had a metric space that gives us this weird topology. How many points are a distance of more than 1 away from pi?

 

[radou]

Yes, you've characterized the open sets. Of course R\{pi} is of the form R\K with K finite.

Yes, but strictly speaking, it doesn't belong to the collection {R\K : K is a finite subset of R and pi is not in K}, but rather to the first collection (it's obtained from it as a union of all of its open singletons).

The point is this. If you have a metric space, then in particular the point pi is some distance from every single point.

Let's look a bit more at pi. Let's suppose we had a metric space that gives us this weird topology. How many points are a distance of more than 1 away from pi?

You mean, in the discrete metric? If so, all points in R\{pi}.

 

[lanedance]

The only open sets containing pi are those whose complement is finite; what does this say about how close almost every point is to pi?

i think this is the key point

 

[Office_Shredder]

Yes, but strictly speaking, it doesn't belong to the collection {R\K : K is a finite subset of R and pi is not in K}, but rather to the first collection (it's obtained from it as a union of all of its open singletons).

Yeah that was stupid on my part

You mean, in the discrete metric? If so, all points in R\{pi}.

No, you don't have the discrete metric. The set R\{pi} has the discrete metric (well, can be given the discrete metric) since it is the discrete topology. Once you thrown pi into the mix you lose that. Remember, open balls around pi must be open sets containing pi

 

[radou]

OK, here's an idea.

Let U be an open set containing pi, so it necessarily must be of the form R\K. Then, for pi, there exists some r > 0 such that the open ball K(pi, r) is in U. Now, such an open ball must again be of the form R\K. The diameter of an open ball of radius r in a metric space is 2r, right? R\K has an infinite diameter, so we arrive at a contradiction.

 

[radou]

Although, when I think twice, I don't think my last post holds, since we don't have any clue about the assumed metric defined on our space, so I can't conclude that the diameter of R\K is infinite.

 

[radou]

I don't know why, but I just can't see the solution to this problem. Hence, I give up.