# A subset of a metric space is closed iff it contains all of its cluster points.

A point, $p$, is defined as a cluster point of a set $S$ if $\forall \epsilon > 0,$ there exists an open ball of radius $\epsilon$ centered at $p$ that contains infinitely many points in $S$.

We want to prove that a subset $S$ of a metric space $(E,d)$ is closed iff $S$ contains all of its cluster points.

I believe I have the forward direction, which seems to be pretty straightforward, but I am having difficulty with the $(\Leftarrow)$ direction. My idea was to somehow use the fact that $S$ contains all of its cluster points to show that every infinite subset of $S$ contains a cluster point. Then we can say $S$ is sequentially compact, therefore it is compact, and hence closed.

I'm just mostly unsure of how to show the first part. Thanks for any comments.

• Your definition of cluster point is strange, to say the least. – José Carlos Santos Sep 20 '17 at 13:49
• What definition of closed set are you using? That statement is sometimes used as a definition of a closed set in a metric space. – TurlocTheRed Oct 13 '18 at 16:55

## 2 Answers

Suppose that $S$ is not closed. The $S\varsubsetneq\overline S$. Take $x\in\overline S\setminus S$. Then $x$ is a cluster point of $S$ that doesn't belong to $S$.

Given the definition $$S\subset E$$ is closed iff the complement of S, $$E\setminus S$$ is open.

A set is Open if all of its points are Interior Points. A point $$x\in S$$ is an Interior point of S if $$\exists \epsilon >0$$ such that if $$d(x,y)<\epsilon$$, then $$y\in S$$. In other words an open ball of radius $$\epsilon$$ centered at $$x$$ contains only elements of $$S$$. So that open ball is itself a subset of $$S$$ completely contained in $$S$$.

Suppose a set $$S$$ contains its cluster points. Let $$x\in S$$ be a cluster point. Then $$\forall \epsilon>0$$, an open ball of radius $$\epsilon$$ centered at $$x$$ contains infinitely many points of $$S$$. Let $$y\in E\setminus S$$ and suppose $$\forall \delta>0$$, an open ball centered at $$y$$ of radius $$\delta$$ contained some point of S. For any such point, reduce $$\delta$$ to be that new point's distance from y. This generates an infinite of points, so $$y$$ is a cluster point of $$S$$, and so must be contained in $$S$$. This contradicts our hypothesis that y is in $$E \setminus S$$. It follows that $$\exists \delta$$ such that an open ball centered at y contains only points in $$E\setminus S$$, so $$y$$ is an interior point of $$E\setminus S$$. By the above, such a $$\delta$$ must exist for every $$y$$, so every $$y$$ is an interior point of $$E\setminus S$$. Since every point of the complement is an interior point, he complement is Open and therefor S is closed.

Other Direction:

Suppose $$S$$ is closed. Then $$E\setminus S$$ is Open. Let $$x\in S$$ be a cluster point of $$S$$. By definition of a cluster point, every open ball centered at $$x$$ must contain an infinite number of points of S. If $$x\in E\setminus S$$, $$x$$ has to be an Interior Point of $$E\setminus S$$ since it is open. But by definition, an interior point must have some open ball centered on it containing no elements of the complement. No cluster point of S can be an interior point of $$E\setminus S$$, so it must not be in $$E \setminus S$$. It is therefor in S, thus S contains all of its cluster points.