If every infinite subset of $S$ has an accumulation point in $S$, then $S$ is bounded.
Proof: Suppose $S$ is unbounded. then , for every $m >0,~~\exists~~x_m \in S$ s.t. $|x_m|>m.$ The collection of $T=\{x_1,x_2,\cdots\}$ is an infinite subset of $S$.
Hence, by the given statement, $T$ has an accumulation point $y$ in $T$.
But, for $m> 1+|y|$, we have $|x_m-y| \geq |x_m|-|y| > m-|y|>1$ contradicting the fact that $y$ is an accumulation point of $T$.
This proves that $T$ is bounded.
Query: What happens when $S$ has only a finite number of unbounded elements in $S$ , that is, if we remove these finite number of unbounded elements from $S$, isn't it possible that $S$ might become a bounded set?
Then, we won't be able to define the set $T$ as collection of infinite elements and neither the accumulation point of $T$ as $y$. Why hasn't the proof accomodated this possibility also?
Thank you for your help..
Best Answer
Let's say that the set $S$ is bounded if we remove a finite number of so-called unbounded elements of it. Then, $S$ is bounded in $B_{r^{*}}(x)$ for some $x \in S$. Now, let $X = \{x_1,x_2,\dots,x_n\}$ be the countable set of all unbounded elements. Let $r = max\{||x - x_i||; x_i \in X\}$, now for every $\rho > r$, the set $S$ is bounded in $B_\rho(x)$.