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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?

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  • $\begingroup$ Isn't this exactly the Bolzano-Weierstrass Theorem? Namely that any bounded sequence has a convergent subsequence? $\endgroup$
    – msteve
    Sep 8, 2014 at 0:48
  • $\begingroup$ Well I dont know, is a homework exercise, and the teacher didnt tell us about it :) $\endgroup$
    – user162343
    Sep 8, 2014 at 0:50
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    $\begingroup$ Once you know that $x_{0}$ is an accumulation point for the bounded set $\{x_{n}: n \in \mathbb{N} \}$, then you can construct a subsequence of $x_{n}$ that converges to $x_{0}$ as follows: For each $k \in \mathbb{N}$, set $A_{k} = (x_{0} - \frac{1}{k}, x_{0} + \frac{1}{k})$ to be the neighborhood centered at $x_{0}$. Since $x_{0}$ is an accumulation point, given any $k \in \mathbb{N}$, there are infinitely many values of $n$ such that $x_{n} \in A_{k}$. Pick $x_{n_{1}} \in A_{1}$. Choose $x_{n_{k}} \in A_{k}$ with $n_{k} > n_{k-1}$.You have it. $\endgroup$ Sep 8, 2014 at 1:10
  • $\begingroup$ Thank you , but, Am I right @akech? or I have to say what you wrote here :) $\endgroup$
    – user162343
    Sep 8, 2014 at 1:19
  • $\begingroup$ Then, Am I right? thank you $\endgroup$
    – user162343
    Sep 8, 2014 at 2:10

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