Real Analysis

[sheriph05]

I am aware a set is Bounded if it has both upper and Lower bounds and i know what a Limit point of a set is but i am finding it hard to show that If $S\subset\mathbb{R}$ be a "bounded infinite set", then $S'\neq\varnothing$ .

[Grigorios Kostakos]

Because $S$ is infinite we can construct a sequence $\{a_n\}_{n\in\mathbb{N}}$ with terms distinct elements of $S$; i.e. a sequence such that \[(\forall\,i,j\in\mathbb{N}) \quad i\neq j \quad \Rightarrow\quad a_i\neq a_j.\] Because $S$ is bounded, $\{a_n\}_{n\in\mathbb{N}}$ is bounded and, by the Bolzano-Weierstrass theorem, $\{a_n\}_{n\in\mathbb{N}}$ has a convergent subsequence $\{a_{k_n}\}_{n\in\mathbb{N}}$. Let $\lim a_{k_n}=a$. Because $\{a_{k_n}\}_{n\in\mathbb{N}}$ is a subsequence of $\{a_n\}_{n\in\mathbb{N}}$ has terms distinct elements of $S$. So, at most only one element of $\{a_{k_n}\}_{n\in\mathbb{N}}$ is equal to $a$. Because $\lim a_{k_n}=a$, we have that for every $\varepsilon>0$ there are infinitely many terms of $\{a_{k_n}\}_{n\in\mathbb{N}}$ in $(a-\varepsilon,a)\cup(a,a+\varepsilon)$. Therefore $a$ is accumulation point (limit point) of $S$, or equivalently, $S'\neq\varnothing$ .