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

Question

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.

Answer 1

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.

Answer 2

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.