# Alternative proof to Theorem 2.37 in Baby Rudin

$$\newcommand{\cl}{\operatorname{cl}}$$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?

It is circular. You are using the fact that $$E$$ has limit points, which is what you are trying to prove.

I agree with the other answer. You're using the fact that $$K$$ is sequentially compact (which is what you are trying to show) when you claim that $$E$$ has limit points. For example, if $$K=\mathbb{R}$$ (as a metric space which is not compact), then you can take $$E=\mathbb{Z}$$ which is closed but has no limit points.

• Thank you. My source of confusion was that, I supposed that infinite subsets of closed sets must have limit points. But this is not the case. Mar 10, 2020 at 22:50

The OP writes

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 $$E'\subseteq cl(E)$$

as a self-contained statement concerning an infinite set $$E$$ in a metric space.

Let the metric space be $$\Bbb R$$ and $$E = \Bbb Z$$.

We have $$cl(E) = E$$ and $$E$$ has no limit points. So the OP's 2) argument falls apart.