How to show that $E$ is closed.

Question

Let $E\subset \mathbb{R}^k$ and suppose that it has the following property

Every infinite subset of $E$ has a limit point in $E$.

I want to show that $E$ is closed.

Efforts:

Idea: I know that idea is to first assume on the contrary that set is not closed. Then find an infinite subset which does not have not a limit point in $E$

Let us assume that $E$ is not closed.(Recall the definition that a set is closed iff it contain all its limit point) So we have a limit point say $x_0$ of $E$ which does not belong to $E$.

Now as $x_0$ is a limit point every nbd of $x_0$ contains a point of $E$ other than $x_0$. In particular take $\delta=1/n$ and choose $x_n$ such that $|x_n-x_0|<1/n$

Let us collect all these points $\{x_n\}$. I have an intutive idea that this is an example of an infinite set of $E$ which does not have the limit in $E$

I am not able to write the ideas rigorously. I would be thankful if somebody can show me the right path.

Answer 1

By your definition of the sequence $(x_{n})$ one can easily see that $(x_{n})$ converges to $x_{0}$ and thus $x_{0}$ is its only limit point which does not lie in $E$.

Answer 2

What you did is fine. Note that $x_0=\lim_{n\to\infty}x_n$. But in a metric space a seuence cannot have more than one limit. So, $x_0$ is the only limit of the sequence $(x_n)_{n\in\mathbb N}$. But if $\{x_n\,|\,n\in\mathbb N\}$ had another limit point $y\in E$, then $y$ would be the limit of a subsequence of $(x_n)_{n\in\mathbb N}$. Now, note that any such subsequence converges to $x_0$ too.

Answer 3

To ensure the $x_n$ are distinct, define $x_n\in B(x_0,\frac1n)\setminus \{x_1,\dots, x_{n-1}\}$. Note that points are closed in a metric space, so taking finitely many points out leaves you with an open set.