Hausdorff spaces and convergence

**Impo** · January 4th 2013, 12:57 PM

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

**Plato** · January 4th 2013, 01:03 PM

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?

**Impo** · January 4th 2013, 01:17 PM

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.

**Plato** · January 4th 2013, 01:53 PM

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 $T_2$ space. Note the capital H.

If $l$ is the limit of $(x_n)$ then any open set $l\in O$ mean all but a finite collect $\{x_1,x_,\cdots,x_n\}$ most belong to $O$ . 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.

**Impo** · January 4th 2013, 02:38 PM

$(2)=>(3)$
If I'm allowed to use the fact that $X$ is Hausdorff then (3) is proved immediately, but I don't think I'm allowed to.

$(3)=>(1)$
Every $T_1$ space is a $T_0$ space, if I can show that $X$ is regular then we're done because every $T_3$ space is Hausdorff. Let $A \subset X$ be a compact subset and suppose $x \notin A$ then there can be found disjoint open neigbourhouds for $A$ and $x$ , but by (3) every compact subspace is closed thus $A$ is closed therefore $X$ is regular.

**jakncoke** · January 4th 2013, 03:03 PM

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

**jakncoke** · January 4th 2013, 06:26 PM

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

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

Now just as we did for a, for b take $\bar{B} = \{N^{b}_i | i \in \mathbb{N} \}$ (its countable basis) and $O^{b}_i = N^{b}_1 \cap ... \cap N^{b}_i | i \in \mathbb{N}$ .
Now for any natural number i, $O^{a}_i \cap O^{b}_i$ 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 $i \in \mathbb{N}$ is $O^{a}_i \cap O^{b}_i = \{ \phi \}$
So if we define a sequence $x_i$ where $x_i \in O^{a}_i \cap O^{b}_i$ for $i \in \mathbb{N}$ . Now i claim $x_i \to a$ . So pick any open neighborhood of a, call it U. since there exists a $j \in \mathbb{N}$ such that $N^{a}_j \subset U$ and since we know $O^{a}_p \subet N^{a}_j \subet U$ for all $p \geq j$ . This shows that for any $x_p$ with $p \geq j$ $x_p \in O^{a}_p$ and thus $x_p \in U$ . Thus $x_i \to a$ .

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

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

**Impo** · January 5th 2013, 04:13 AM

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.

**Plato** · January 5th 2013, 06:13 AM

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:

Originally Posted by Impo

$(2)=>(3)$
If I'm allowed to use the fact that $X$ 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.

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