Hausdorff spaces and convergence

Hi Everyone,

I'm new here and I'm dealing with some topology questions.

Suppose that (X,T) is a C1 space then the following statements are equivalent:

(1) (X,T) is a hausdorff space

(2) Every convergence sequence has a unique limit

(3) (X,T) is a T1-space and every compact subset is closed.

I know this is not a homework service but honestly I have no idea where to start. In fact, in topology I always described convergence using filters and in this question I have to prove something about sequences so I'm confused.

Can someone give me a hint?

Thanks in advance!

Cheers Impo

Re: Hausdorff spaces and convergence

Please tell us what "Suppose that (X,T) is a C1 space" means.

C1 is not a common notation in topology. Does it mean *first countable*?

Re: Hausdorff spaces and convergence

Quote:

Originally Posted by

**Plato** Please tell us what "Suppose that (X,T) is a C1 space" means.

C1 is not a common notation in topology. Does it mean *first countable*?

Indeed.

Re: Hausdorff spaces and convergence

Re: Hausdorff spaces and convergence

If I'm allowed to use the fact that is Hausdorff then (3) is proved immediately, but I don't think I'm allowed to.

Every space is a space, if I can show that is regular then we're done because every space is Hausdorff. Let be a compact subset and suppose then there can be found disjoint open neigbourhouds for and , but by (3) every compact subspace is closed thus is closed therefore is regular.

Re: Hausdorff spaces and convergence

Well you are supposed to prove IFF implications. and proving these two things is akin to showing all three statements are equivalent. Plato already gave you now prove

Re: Hausdorff spaces and convergence

So let me give you some help for

Now a (X,T) being a C1 space means that every point has a countable neighborhood basis. (Meaning a sequence of open neighborhoods such that for any open neighborhood U of a, there exists an element in the sequence which is contained in U.

Now if a, b were points in the topology, such that . Since C1, a has a countable neighborhood basis, call it where is an open neighborhood of a. So for any open neighborhood U of a, there exists an such that . Now since the intersection of any finite number of open sets is open. . For any let Note that for all . Note that since for . This should remind of you the definition for topological convergence. That is, a sequence if for any open neighborhood of a, we can find a such that implies .

Now just as we did for a, for b take (its countable basis) and .

Now for any natural number i, is an open set (containing a, and containing b), since it is the finite intersection of 2 open sets.

Now if we assume a and b have no disjoint open sets containing a and b. Then for no is

So if we define a sequence where for . Now i claim . So pick any open neighborhood of a, call it U. since there exists a such that and since we know for all . This shows that for any with and thus . Thus .

Now since we assumed, for , then . I claim . Now again pick any open neighborhood of b, call if U. then there exists a such that which is then for all . so this shows that for all , thus .

Since we assumed every convergent sequence has a unique limit, which is a contradiction. Thus there be an open set containing a which is disjoint with an open set containing b. Thus hausdrauff.

Re: Hausdorff spaces and convergence

Hi jakncoke, thanks for your answer. I wanted to prove the statements are equivalent by proving (1)=>(2)=>(3)=>(1). I'm only missing (2)=>(3) for now.

Re: Hausdorff spaces and convergence

Quote:

Originally Posted by

**Impo** I wanted to prove the statements are equivalent by proving (1)=>(2)=>(3)=>(1). I'm only missing (2)=>(3) for now.

I commend you for wanting to show equivalence that way. But as you have already noted:

Quote:

Originally Posted by

**Impo**
If I'm allowed to use the fact that

is Hausdorff then (3) is proved immediately, but I don't think I'm allowed to.

I think that I would use some sort of argument such as jakncoke used in reply #7 to show the space is Hausdorff. Then as you note it is an easy to precede.