Let $(X, d)$ be a metric space and $(x_n)_n$ a convergent sequence in $X$ with $x := \lim_{n \to \infty} x_n$. Denote $A := \{x_n : n \in \mathbb N \}$.
Then it is pretty clear that $\overline{A} = \{x\} \cup A$.
"$\supseteq$": This is clear since $A \subseteq \overline A$ by definition and $x \in \overline A$ since $x$ is a limit point of $A$.
"$\subseteq$": This direction is quite tricky.
I find it hard to give an elegant proof that only shows this inclusion without employing compactness. More precisely, it is rather easy to show that $\{x\} \cup A$ is compact and thus closed. Hence $\overline{A} \subseteq \{x\} \cup A$ since $\overline{A}$ is the smallest closed set that contains $A$.
But that is not what I am after: Suppose that $\overline A \setminus A \neq \emptyset$ and let $y \in \overline A \setminus A$. Can I show that in this case $y = x$ holds just by using the definition of the closure? Since $y \in \overline A$ I would know that for each $\varepsilon > 0$ there is $z \in A$ such that $d(y, z) \leq \varepsilon$. Hence
$$d(x,y) \leq d(x,z) + d(z, y) \leq d(x,z) + \varepsilon.$$
But how can I get $d(x,z) \leq \varepsilon$? Anyhow I need to get an element $z \in A$ that is simultanously close to $x$ and $y$ but I do not quite see how to achieve that. Any thoughts?
Best Answer
For each $k$ there exists (infinitely many) $n_k$ such that $d(y,x_{n_k}) <\frac 1 k$. We may assume that $n_k$ is increasing. This implies that $x_{n_k} \to y$. Together with $x_n \to x$ we get $y=x$.