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$\{x_k\}$ is a sequence of real numbers and let $\Omega$ be the set of all distinct points in $\{x_k\}$. Prove that if $\Omega$ has a limit point $x$, then $\{x_k\}$ has a sub-sequence converging to $x$.

At first glance it seems easy but somehow I got a problem. Although the definition of limit point indicates we can construct a sequence (let's denote it by $\{y_j\}$) with distinct elements out of $\Omega$ and make it converge to $x$, but how do we decently show $\{y_j\}$ is a subsequence of $\{x_k\}$?

I also tried to prove by contradiction by supposing there exists no such subsequence, then for any subsequence $\{x_{k_l}\}$ we have $\exists \varepsilon >0 \forall K \exists k_l>K$ such that $\left| {{x_{{k_l}}} - x} \right| \ge \varepsilon $. But I got stuck and don't know how to proceed to show there is contradiction with the existence of limit point.

Hope someone can help! Thank you.

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Say $x^*$ is a limit point of $\Omega$. That is, for each $\epsilon > 0$ there exists some $x\in\Omega\setminus\{ x^* \}$ with $|x-x^*| < \epsilon.$ Thus, the set $$ X_{\epsilon} = \{ x\in\Omega\setminus\{ x^* \} : |x-x^*| < \epsilon \} $$ infinite. Now, you can define the indices for the subsequence recursively: $$ k_1 = \min\{ k \ge 1 : x_k \in X_1 \} $$ and for $n \ge 1$: $$ k_{n+1} = \min\{ k \ge k_n : x_k \in X_{1/(n+1)} \}. $$ Then, $(x_{k_n})_{n\ge 1}$ is a subsequence and converges to $x^*$ as $$ |x_{k_n} - x^* | < \frac{1}{n}. $$

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  • $\begingroup$ I got it. Thanks! $\endgroup$ – Tony Jul 5 '15 at 12:37

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