Prove that a set $E \subset R^n$ is closed and bounded if and only if every infinite subset of $E$ has a point of accumulation (aka limit point) that belongs to E
Hi all,
Working on an exercise from Thomson, Bruckner & Bruckner's Elementary Real Analysis and been stuck on this one for quite some time. Here is what I have so far.
$\Leftarrow $
Suppose every infinite subset $E_k$ of $E$ has a point of accumulation that belongs to $E$.
We wish to show that $E$ is closed and bounded.
First, suppose that $E$ is not closed.
Then for all limit points, $x_0$ of $E_k$, $x_0 \in E_k \land x_0 \notin E$.
But by assumption, every infinite subset $E_k \subset E$ contains a point of accumulation of $E$, so this is a contradiction.
Therefore $E$ is closed.
$\Rightarrow$
Now, suppose $E$ is not bounded. . .
It's this right implication that I am stuck on. I can't find any justification for the argument that I want to make, which is that if any infinite subset of E contains limit points of E, then E must be bounded. This is based off the second argument from this question's top answer, from Chocosup. Every infinite subset of E in R having a limit point in E implies E is closed and bounded
Any help would be very much appreciated!
Best Answer
If $E$ is unbounded, for each $k \in \mathbb N$ choose $a_k \in E$ with $\|a_k\| \ge k$. Then $A:=\{a_1,a_2,\dots\}$ has no accumulation point.
Suppose $a \in R^n$. I claim that $B_1(a) \cap A$ is finite, so $a$ is not an accumulation point of $A$. Let $L = \|a\|$. Note if $k \ge L+2$ then $\|a_k\| \ge k \ge L+2$ and so by the triangle inequality $\|a_k-a\| \ge L+2-L > 1$, so $a_k \notin B_1(a)$.
Notation $B_r(a) = \{x \in R^n : \|a-x\| < r\}$.