# Difficulty in understanding details behind Rudin's $\;$ '...clearly has no limit points in $R^k$'

I recently begun to read Walter Rudin magnum upos "Principles of Mathematical Analysis" and i'm having a little trouble in understanding the proof of the the following statement:

2.41 Theorem: if a set $E$ in $R^k$ has one of the following three properties then it has the other two:

(a) $E$ is closed and bounded;

(b) $E$ is compact;

(c) Every infinite subset of $E$ has a limit point in $E$.

He proves that (a) implies (b) and that (b) implies (c), remaining only to show that (c) implies (a). He started as follow:

If$E$ is not bounded, then $E$ contains points $x_n$ with

$|x_n| > n ( n = 1,2,3...)$

The set $S$ consisting of these points $x_n$ in infinite and clearly has no limit points in $R^k...$

My question is why this set $S$ has no limit points in $R^k$?

• Your condition (c) makes no sense. It must be " Every infinite subset of $\;E\;$ has a limit point in $\;E\;$" Feb 19, 2018 at 12:59
• Sorry. I've missed that. Fixed Feb 19, 2018 at 13:01
• I know. Please do edit your question. Feb 19, 2018 at 13:02

We have our set $S = \{x_n \;|\; n \ge 1\}$ with $|x_n| \gt n$.
Let $x$ be any point in $\Bbb R^k$ . Then $x$ is not a limit point of the set $S$.
To show this, first find an integer $N \ge 1$ so that $x$ is an interior point of the closed ball $B_N$ of radius $N$ about the origin (zero coordinates). There are at most $N - 1$ points of $S$ that can also be contained in this ball. We can form an open ball about $x$ contained in $B_N$ that excludes any one of these points, except perhaps for $x$ itself. But then since the finite intersection of open sets is open, we can find an open set containing $x$ that excludes all points of $S$, except, of course, for $x$ itself if it is in $S$.
Answer to question: because either $\;\lim\limits_{n\to\infty} x_n\;$ doesn't exist or else it is $\;\pm\infty\;$ (one of the two). Either way, the limit is not in $\;\Bbb R^k\;$ .