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.
Best Answer
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.