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

Quote:

Originally Posted by

**Impo** 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

Note that a Hausdorff space is a $\displaystyle T_2$ space. Note the capital H.

If $\displaystyle l$ is the limit of $\displaystyle (x_n)$ then any open set $\displaystyle l\in O$ mean all but a finite collect $\displaystyle \{x_1,x_,\cdots,x_n\}$ most belong to $\displaystyle O$. If you have two limits then in a Hausdorff there are disjoint open sets that separate those limits. That is impossible.

Now you post some work.

Re: Hausdorff spaces and convergence

$\displaystyle (2)=>(3)$

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

$\displaystyle (3)=>(1)$

Every $\displaystyle T_1$ space is a $\displaystyle T_0$ space, if I can show that $\displaystyle X$ is regular then we're done because every $\displaystyle T_3$ space is Hausdorff. Let $\displaystyle A \subset X $ be a compact subset and suppose $\displaystyle x \notin A$ then there can be found disjoint open neigbourhouds for $\displaystyle A$ and $\displaystyle x$, but by (3) every compact subspace is closed thus $\displaystyle A$ is closed therefore $\displaystyle X$ is regular.

Re: Hausdorff spaces and convergence

Well you are supposed to prove IFF implications. $\displaystyle (1) \text{if and only if} (2) $ and $\displaystyle (2) \text{if and only if} (3) $ proving these two things is akin to showing all three statements are equivalent. Plato already gave you $\displaystyle (1) \to (2) $ now prove $\displaystyle (2) \to (1) $

Re: Hausdorff spaces and convergence

So let me give you some help for $\displaystyle (2) \to (1) $

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 $\displaystyle a \not = b $. Since C1, a has a countable neighborhood basis, call it $\displaystyle \bar{A} = \{ N^{a}_i | i \in \mathbb{N} \} $ where $\displaystyle N^{a}_i $ is an open neighborhood of a. So for any open neighborhood U of a, there exists an $\displaystyle i \in \mathbb{N} $ such that $\displaystyle N^{a}_i \subset U $. Now since the intersection of any finite number of open sets is open. $\displaystyle N_1 \cap N_2 \cap ... \cap N_i \subset N_i \subset U $. For any $\displaystyle i \in \mathbb{N} $ let $\displaystyle O^{a}_i = N^{a}_1 \cap ... \cap N^{a}_i $ Note that $\displaystyle O^{a}_j \subset N^{a}_i $ for all $\displaystyle j \geq i $. Note that since $\displaystyle O^{a}_j \subset N^{a}_i \subset U $ for $\displaystyle j \geq i $. This should remind of you the definition for topological convergence. That is, a sequence $\displaystyle x_n \to a $ if for any open neighborhood of a, we can find a $\displaystyle j \in \mathbb{N} $ such that $\displaystyle p \geq j $ implies $\displaystyle x_p \in U $.

Now just as we did for a, for b take $\displaystyle \bar{B} = \{N^{b}_i | i \in \mathbb{N} \} $ (its countable basis) and $\displaystyle O^{b}_i = N^{b}_1 \cap ... \cap N^{b}_i | i \in \mathbb{N} $.

Now for any natural number i, $\displaystyle O^{a}_i \cap O^{b}_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 $\displaystyle i \in \mathbb{N} $ is $\displaystyle O^{a}_i \cap O^{b}_i = \{ \phi \} $

So if we define a sequence $\displaystyle x_i $ where $\displaystyle x_i \in O^{a}_i \cap O^{b}_i $ for $\displaystyle i \in \mathbb{N} $. Now i claim $\displaystyle x_i \to a $. So pick any open neighborhood of a, call it U. since there exists a $\displaystyle j \in \mathbb{N} $ such that $\displaystyle N^{a}_j \subset U $ and since we know $\displaystyle O^{a}_p \subet N^{a}_j \subet U $ for all $\displaystyle p \geq j $. This shows that for any $\displaystyle x_p $ with $\displaystyle p \geq j $ $\displaystyle x_p \in O^{a}_p $ and thus $\displaystyle x_p \in U $. Thus $\displaystyle x_i \to a $.

Now since we assumed, $\displaystyle x_i \in O^{a}_i \cap O^{b}_i $ for $\displaystyle i \in \mathbb{N} $, then $\displaystyle x_i \in O^{b}_i $. I claim $\displaystyle x_i \to b $. Now again pick any open neighborhood of b, call if U. then there exists a $\displaystyle j \in \mathbb{N} $ such that $\displaystyle N^{b}_j \subset U $ which is then $\displaystyle O^{b}_p \subset N^{b}_j \subet U $ for all $\displaystyle p \geq j $. so this shows that for all $\displaystyle p \geq j $ $\displaystyle x_p \in O^{b}_p \subset U $, thus $\displaystyle x_i \to b $.

Since we assumed every convergent sequence has a unique limit, $\displaystyle a = b $ 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** $\displaystyle (2)=>(3)$

If I'm allowed to use the fact that $\displaystyle X$ 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.