$\newcommand{\cl}{\operatorname{cl}}$In Rudin's Principles of Mathematical Analysis, Theorem 2.37 says the following:
If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.
I think this statement can be proved even easier than given in the book, if we rely on some Theorems proven earlier in the book:
1) Theorem 2.34 says that compact sets of metric spaces are closed. Therefore $K$ is closed.
2) The closure of $E$, $\cl(E)$ is by definition is closed, and since $E$ is infinite, the set of limit points of $E$, $E'$ is nonempty and again by definiton $E'\subseteq \cl(E)$
3) Theorem 2.27 c) says that $\cl(E)\subseteq F$ for every closed set $F$ such that $E\subseteq F$
So $E\subseteq \cl(E) \subseteq K$, and we are done. Are there any flaws in this logic, or this is indeed an alternative proof?
Best Answer
It is circular. You are using the fact that $E$ has limit points, which is what you are trying to prove.