# Cauchy sequence with $\{ x_n : n \in \mathbb{N} \}$ not closed converges

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.

• What is $X$ ? Where the $x_n$ are from ? – Fred May 14 at 9:35
• ... and what's $S(y,\varepsilon)$? – Saucy O'Path May 14 at 9:35
• $X$ is a metric space and $S(y, \varepsilon)$ is an open sphere with center $y$ and radius $\varepsilon$. – LoneBone May 14 at 9:38

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.
• Why the fact that $x$ does not belong to the sequence ensures that it's the limit of some subsequence of $x_n$? – LoneBone May 14 at 9:42
• 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. – José Carlos Santos May 14 at 9:45
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.