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A subequence of a sequence $(x_n)_{n\ge 1}$ is a sequence $ (x_{n_1}, x_{n_2},x_{n_3},...$) where $n_1,n_2,n_3,... \in \Bbb N$ with $n_1\lt n_2\lt n_3\lt ...$

Let $(X,d)$ be a metric space and let $x\in X$. Prove that if $x\in X$ and sequence $(x_n) \in X^{\Bbb N}$ converges to $x$ in $(X,d)$ , then so does every subsequence of $(x_n)$.

Defnition: Let $(X, d)$ be a metric space, let $(x_n)\in X^{\Bbb N}$ and let $x\in X$. We say that $(x_n)$ converges to $x$ in $(X, d)$, if $\forall B\in \mathscr B(x) : x_n\in B $ eventually

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  • $\begingroup$ What have you tried? What does it mean for the sequence to converge to $x$? What would it mean for a subsequence to converge to $x$? $\endgroup$ – Hagen von Eitzen Apr 7 '13 at 18:31
  • $\begingroup$ $\mathscr B(x)$ is the set of all balls centerd at $x$ with raduis r $\endgroup$ – Jhwana Apr 7 '13 at 18:36
  • $\begingroup$ @HagenvonEitzen : I used the above definition but I couldn't deduce that every subsequence is converges to the same limit. $\endgroup$ – Jhwana Apr 7 '13 at 18:40
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Suppose that for each $\epsilon >0$ we can find $N$ which makes $d(x_n,x)<\epsilon$ by taking $n\geq N$. Consider a subsequence $\langle x_{n_j}\rangle$. Since $n_1<n_2<\dots$, there must exist a $J$ for which $n_j\geq N$ whenever $j\geq J$. In such as case, what can you say about $$d(x_{n_j},x)\text{ ? }$$

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  • $\begingroup$ $d(x_{n_j},x)\text{ }< \epsilon$... so $(x_{n_j})$ converges to $x$... thank u $\endgroup$ – Jhwana Apr 7 '13 at 19:12
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Note that "$x_n\in\mathscr B$ eventually" means the same as "There are at most finitely many $n$ with $x_n\notin \mathscr B$". If this holds for the full sequence, it must even more so hold for a subsequence.

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