Existence of a limit point implies existence of inifintely many limit points?

[julypraise]

Homework Statement

Prove the following statement is true or not:

the statement:
Let $(X,d)$ be a non-empty metric space and $A$ is a non-empty subset of $X$. Then if $A'$ is not empty, then $A'$ is infinte.

Homework Equations

Definition of limit point and its negation.

The Attempt at a Solution

I tried to prove by contradiction in this way: (to prove the statement is true)
Suppose $p_{1},\dots,p_{N}$ are limit points of $A$. Since $A$ has a limit point, it is infinte. Now observe that if $p'\in A,p\notin\{p_{1},\dots,p_{N}\}$ then $p'$ is not a limit point of $A$, i.e., $A\cap B(p';r_{p'})=\{p'\}$ for some $r_{p'}>0$. Thus $A$ has a kind of gap inside it.

But from here, I cannot go further. It almost seems that $A'$ finite is consistent.

 

[Dick]

But from here, I cannot go further. It almost seems that $A'$ finite is consistent.

Go with that. I can think of a metric space with one limit point. Can you?

 

[julypraise]

Go with that. I can think of a metric space with one limit point. Can you?

Sorry, I tried but couldn't think of one. Could you give me at least a hint? Like, what kind of metric space is it with what kind metric??

 

[HallsofIvy]

Well, one obvious way to avoid "an infinite number of limit points" is to start with a space that only contains a finite number of points!

 

[julypraise]

Well, one obvious way to avoid "an infinite number of limit points" is to start with a space that only contains a finite number of points!

Humm.. But, you know, if a set has only finite points then there is no limit point of this set for whatever metric space is talked about. Isn't it?

 

[Dick]

Sorry, I tried but couldn't think of one. Could you give me at least a hint? Like, what kind of metric space is it with what kind metric??

Nothing fancy. Pick a subset of the real numbers. They should be discrete but approach a single limit point. Like 0.

 

[julypraise]

Wow, this one right? $\bigcup_{n=1}^{\infty}\{-\frac{1}{n},\frac{1}{n}\}$.

 

[Dick]

Wow, this one right? $\bigcup_{n=1}^{\infty}\{-\frac{1}{n},\frac{1}{n}\}$.

Looks generally ok to me. But I think you want to include 0 in your subset. What do you think?

 

[julypraise]

Looks generally ok to me. But I think you want to include 0 in your subset. What do you think?

Ah... I think it's okay to disprove the statement. I showed that A' is not empty but finite anyway. Thanks!