# Close of a Singleton in R1

1. Mar 10, 2013

### gajohnson

Let me add some clarification here, since I realized that my notation might be confusing. By $\left\{x\right\}'$ below, I mean the set of all limit points of $\left\{x\right\}$. Hope this helps anyone trying to figure out what I mean, thanks!

The title says R1, but it's actually just a metric space $X$, my apologies.

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

NA

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!

Last edited: Mar 10, 2013