Does a limit point always exist?

Question

My instructor proves a statement during a lecture: if a set $E$ is open in a metric space $X$, then $E^c$ is closed.

In his proof, he writes: Suppose $E$ is open. We want to show that $E^c$ is closed (contains all its limit points). Let $x$ be a limit point of $E^c$. Then......

My question is how to know there exists a limit point in $E^c$? Thanks!

Answer 1

The proof is correct, but does not explicitly state the situation where the set of limit points is empty, starting off with such a limit point. A reformulation, just to make it clearer to the OP, would be : Let $S$ be the set of limit points of $E^c$. We want to show that $S$ is a subset of $E^c$.

If $S$ is empty, then the empty set is contained in every set, so $S \subset E^c$ and $E^c$ contains all its limit points. (Stating the empty case explicitly)

If not, then let $x \in S$ be a limit point of $E^c$. Then , continue as in the stated proof.

So $x \in E^c$. Consequently $S \subset E^c$ and hence $E^c$ is closed.

Answer 2

You want to show that $(E^c)' \subseteq E^c$. Where for a set $A$, $A'$ denotes the set of its limit points.

To show an inclusion $A \subseteq B$, we usually let $x$ be an arbitary element of $A$ and show it must be in $B$. If there are no elements in $A$ this is vacuously true, so that's no problem.

So in the above vein we thus start by letting $x$ be an arbitrary point of $(E^c)'$. If there is no such point we're done anyway. ("nothing to be done")