Heine Borel question Rudin

Question

In Baby Rudin, there is proven that if every infinite subset of a set $E$ has a limit point, then the set is closed:

See Baby Rudin theorem 2.41 for more details and a proof.

I have a quick question. In the part where he proves that E is closed, he mentions that $x_0$ is a limit point of S. I can see that this is true, but I don't understand how it is relevant for the proof. It only seems to matter that $x_0 \notin E$, so it can't be a limit point of $S$ in $E$ regardless whether or not it actually is a limit point.

Am I missing something or am I correct?

Answer 1

You are right. I guess Rudin wanted to first point out all the limit points of $S$ and after seeing no limit point of $S$ in $E$, he then made the contradict argument.

Answer 2

Rudin is proving by contrapositive here, suppose $E$ is not closed, then there has to be a limit point $x_0$ of $E$ s.t. $x_0 \notin E$, note that $E$ can also be closed when having no limit points at all, i.e. vacuously , and that’s why we need to construct $S \subset E$ s.t. $S$ has a limit point $x_0 \notin S$.

Therefore we can’t say that 'It only seems to matter that $x_0 \notin E$ 'etc..