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Suppose $(x_n)_{n \in \mathbb{N}}$ is a Cauchy sequence and $A = \{ x_n : n \in \mathbb{N} \}$ not closed. Show that there exists $x \in X$ such that $x_n \longrightarrow x$.

Since $A$ is not closed: $$ \forall y\in A , \exists \varepsilon > 0 s.t. S(y,\varepsilon) \subset A $$ and since $x_n$ is Cauchy: $$ \forall \varepsilon > 0 ,\exists n_0 \in \mathbb{N} s.t. \forall n,m \geq n_0 : \rho (x_n,x_m) < \varepsilon $$ But I can't see a way to connect these two to prove convergence. Some starting hints would be greatly appreciated.

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  • $\begingroup$ What is $X$ ? Where the $x_n$ are from ? $\endgroup$ – Fred May 14 at 9:35
  • $\begingroup$ ... and what's $S(y,\varepsilon)$? $\endgroup$ – Saucy O'Path May 14 at 9:35
  • $\begingroup$ $X$ is a metric space and $S(y, \varepsilon)$ is an open sphere with center $y$ and radius $\varepsilon$. $\endgroup$ – LoneBone May 14 at 9:38
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Take $x\in\overline{\{x_n\,|\,n\in\mathbb N\}}\setminus\{x_n\,|\,n\in\mathbb N\}$. Then there is some sequence of terms of the sequence $(x_n)_{n\in\mathbb N}$ converging to $x$. In other words (since $x$ itself does not belong to the sequence), $x$ is the limit of some subsequence of the sequence $(x_n)_{n\in\mathbb N}$. But whenever a Cauchy sequence has a convergent subsequence, the whole sequence converges to the limit of that subsequence.

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  • $\begingroup$ +1. A quick answer. $\endgroup$ – Qurultay May 14 at 9:41
  • $\begingroup$ Why the fact that $x$ does not belong to the sequence ensures that it's the limit of some subsequence of $x_n$? $\endgroup$ – LoneBone May 14 at 9:42
  • $\begingroup$ There is a sequence $(n_k)_{k\in\mathbb N}$ such that $\lim_{k\to\infty}x_{n_k}=x$. Is it possible that the set $\{n_k\,|\,k\in\mathbb N\}$ is finite? No, because then there would be a $N\in\mathbb N$ such that $n_k=N$ infinitely many times and we would have to have $x_N=x$. But $x$ does not belong to the sequence. $\endgroup$ – José Carlos Santos May 14 at 9:45
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Not closed means there is a sequence from the set which converges to a point not in the set. Let $(y_n)$ be a sequence from $\{x_n:n \geq 1\}$ such that $y_n \to y$ and $y\neq x_n$ for any $n$. For each $n$ we can write $y_n=x_{k_n}$ and this sequence cannot be equal to $x$ after some stage because $y\neq x_n$ for any $n$. Hence we can find a subsequence of $\{x_n\}$ which is convergent. It is well known fact that if a subsequence of a Cauchy sequence converges then the whole sequence converges.

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