# Hausdorff spaces and convergence

• Jan 4th 2013, 12:57 PM
Impo
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?

Cheers Impo
• Jan 4th 2013, 01:03 PM
Plato
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?
• Jan 4th 2013, 01:17 PM
Impo
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.
• Jan 4th 2013, 01:53 PM
Plato
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.
• Jan 4th 2013, 02:38 PM
Impo
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.
• Jan 4th 2013, 03:03 PM
jakncoke
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 $\displaystyle (2) \to (1)$
• Jan 4th 2013, 06:26 PM
jakncoke
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.
• Jan 5th 2013, 04:13 AM
Impo
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.
• Jan 5th 2013, 06:13 AM
Plato
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.