I need to prove the following:
Let $\{a_{n}\}$ be a bounded sequence of real numbers.Prove that $\{a_{n}\}$ has a convergent subsequence.(Hint: You may want to use the Bolzano-Weierstrass Theorem)
My attempt:
Since $\{a_{n}\}$ is a set that is bounded, non-empty, and infinite, it has by the Bolzano-Weierstrass Theorem an accmulation point, lets call it $x_{0}$. We have the next result: Let $A \subset \mathbb{R}$, $a_{0}$ is an accumulation point of $A$ iff exists a sequence $\{x_{n}\}$ in $A$ such that $\{x_{n}\} \rightarrow a_{0}$ with $x_{0}$ diferent from $a_{0}$ for all n.
Then $A$ is our sequence $\{a_{n}\}$ and $a_{0}$ is the accumulation point $x_{0}$, therefore by the above result exists a subsequence $\{b_{n}\}$ with $b_{n}$ diferent from $x_{0}$ in $\{a_{n}\}$ s.t. converges to $x_{0}$.
Am I right?