Let $(X,d)$ be a metric space, and let $\{x_n\}_{n \in \mathbb{N}}$ be a sequence in $X$. Assume that every subsequence of $\{x_n\}_{n \in \mathbb{N}}$ has a sub-subsequence that converges to the same $x^*\in X$. Show that the origininal sequence also converges to this $x^*$.

This is the solution that the professor provided and I can't understand the second part:

Assume that the sequence $\{x_n\}_{n \in \mathbb{N}}$ does not converge to $x^*$. There exists some $\epsilon_0 >0$ such that, for all $n \in \mathbb{N}$, there exists some natural $M(n) \geq n$ such that $d(x_{M(n)},x^*) \geq \epsilon_0$. Construct the subsequence $\{x_{k(n)}\}_{n \in \mathbb{N}}$ by choosing $k(1)=M(1)$ and $k(n)=\max\{M(n),M(k(n-1)+1)\}$ for $n \geq2$. Thus subsequence has no sub-subsequence that converges to $x^*$.

I don't understand the subsequence is constructed in such a why it's true that the subsequence has no sub-subsequence that converges to $x^*$.

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    $\begingroup$ The subsequence is constructed in such a way that $d(x_{k(n)},x^*)\ge \epsilon_0$ for all $n$. This property will then also hold for any sub-subsequence, so there can't be convergence to $x^*$. $\endgroup$ – Sofia Oct 25 '14 at 15:30
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    $\begingroup$ By construction, every term of the subsequence $(x_{k(n)})$ has a distance of at least $\epsilon_0$ from $x^\ast$. $\endgroup$ – Daniel Fischer Oct 25 '14 at 15:30
  • $\begingroup$ Could you expand on that? $\endgroup$ – Rby Oct 25 '14 at 16:11

If $x_{k(n)}$ had a subsequence converging to $x^*$, then the inequality $d(x_{k(n)},x^*)<\epsilon_0$ would hold for all sufficiently large $n$. (Why? Review the definition of limit). But this is not the case.

  • $\begingroup$ The part I'm not getting is the construction of the subsequence, not the definition, I don't understand why the subsequence is constructed that way. $\endgroup$ – Rby Oct 26 '14 at 15:01

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