# Proving {x} is a closed set in a metric space

1. Mar 12, 2013

### gajohnson

Hi everyone, I posted this a couple days ago and didn't get a response, so I thought I'd try again. Let me know if something about this is confusing. Thanks!

1. The problem statement, all variables and given/known data

Let $X$ be a metric space and let $x\in{X}$ be any point. Prove that the set $\left\{x\right\}$ is closed in $X$

2. Relevant equations

I am defining $\left\{x\right\}'$ as the set of all limit points of $\left\{x\right\}$.

3. The attempt at a solution

Ok, so I realize I can prove this by showing that the complement is open (simple). However, I have a question about the validity of an alternative approach:

It is clear that $\left\{x\right\}$ has no limit points, thus $\left\{x\right\}' =$empty set. Now, the empty set is a subset of every set, thus $\left\{x\right\}'\subset{\left\{x\right\}}$ and so $\left\{x\right\}$ contains all of its limit points. Then $\left\{x\right\}$ is closed.

Thoughts? Does this work? If not, what am I missing? Thanks!

2. Mar 12, 2013

### jbunniii

Yes, this proof is fine.

3. Mar 12, 2013

### gajohnson

Ok, thanks for validation. I had used this proof on an exam and it was marked almost entirely wrong, which was confusing to me. I'll question the professor about it. Thanks again!

4. Mar 12, 2013

### jbunniii

The proof is correct, but perhaps he wanted more detail. Definitely check with him - only he can tell you what the problem is.

Maybe he didn't like the key statement: "it is clear that {x} has no limit points" - it's clear to me, but perhaps he wanted more detail to be sure why it was clear to you.

Also, if a closed set is defined as the complement of an open set, then it's a theorem (perhaps requiring proof on an exam) that this is equivalent in a metric space to "the set contains all of its limit points."

5. Mar 12, 2013

### gajohnson

Ah yes, actually on the proof on the exam I showed clearly that {x} had no limit points--I just glossed over it here since it seemed like it would not be a possible point of contention in the proof. We were also able to freely use theorems covered in Rudin, so the rest should be fine.

Thanks again!