Let $E$ be a subset of $R^{k}$, with the Euclidean metric on it. Then $1.$ implies $2.$, where
$1.$ Every infinite subset of $E$ has a limit point in $E$
$2.$ $E$ is closed.
Proof Suppose by contradiction that $E$ is not closed. Since by definition a subset of a metric space is said to be closed if it contains all its limit points, we are assuming there exists a point $x_0$ of $\mathbb{R}^k$ which is a limit point of $E$ and which lies in the complement of $E$ with respect to $\mathbb{R}^k$. For every $n\in\mathbb{N}$, there are points $x_n\in E$ such that $|x_n-x_0|<\frac{1}{n}$. Let $S$ be the set of these points. Then $S$ is infinite (otherwise $|x_n-x_0|$ would have a constant positive value, for infinitely many $n$)....
I can't understand the "otherwise" argument. Since He said $S$ is infinite, I suppose that "otherwise" means that $S$ is by contradiction assumed to be finite. But then how can we consider infinitely many indices $n$, so that $x_n$ is a point of $S$ for each $n\in\mathbb{N}$? I think the only way is to "count" some of them infinitely many times. But then how can the condition $|x_n-x_0|<\frac{1}{n}$ hold for all of them, considering that $\frac{1}{n}$ converges to $0$, the only point $x_n$ with $|x_n-x_0|=0$ is $x_0$ and $x_0\notin S$?
Could you please expand Rudin's argument so to make it more understandable to me? Thank you.
To elaborate on the "otherwise", let us look at the negation.
That is, let us ask ourselves what would happen if that were not true.
Suppose that $S$ were finite.
Note that $S$ was constructed in a way such that we have the following property:
Now, since $S$ is finite, this means that not all $x_n$ can be distinct. In fact, it would mean that there's some $x' \in S$ such that $$x' = x_n\quad\text{for infinitely many } n \in \Bbb N.$$
(Why? If that didn't happen, then $\Bbb N$ could be written as a finite union of finite sets.)
Let $A$ be the collection of all such $n$ that satisfy the above condition. This is an infinite set. What Rudin is saying is that
$$|x_n - x_0| \text{ is consant for all } n \in A.$$ And that is simply because $|x_n - x_0| = |x' - x_0|$ for all $n \in A$.
This is why $|x_n - x_0|$ is constant for infinitely many $n$.
Moreover, this constant is positive because $x_0 \notin S \ni x'$ and thus, $x' \neq x_0$.
The contradiction that he would derive from this is that a fixed positive number $\delta = |x' - x_0|$ is less than $1/n$ for infinitely many $n$. (Since it's infinitely many, you choose an $n > 1/\delta$ and arrive at the contradiction.)