# Closure of convergent sequence

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?

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 $$y=x$$.

• You are right. But to see the "you may assume"-part you need to distinguish some two different cases and I wondered if you could surpass this argument. But honestly, I don't see how you could. – Adriano Oct 2 '19 at 9:11
• @Adriano If $y \in \overline {A} \setminus A$ then there exist infinitely many $n$'s with $d(y,x_n) <\frac 1 k$. So you can make $n_1<n_2<...$. – Kavi Rama Murthy Oct 2 '19 at 9:16

Note that in a metric space,we have that $$x \in \bar{B}$$ iff exists $$x_n \in B$$ such that $$x_n \to x$$

Now let $$y \in A\cup\{x\}$$

If $$y=x_m$$ for some $$m \in \Bbb{N}$$ then take the constant sequence. $$y_n=x_m,\forall n \in \Bbb{N}$$

If $$y=x$$ then $$x_n \in A$$ and $$x_n \to x$$

So $$\bar{A}=A \cup \{x\}$$

If $$y \notin A \cup \{x\}$$ then $$y \neq x$$ and $$y\neq x_n,\forall n \in \Bbb{N}$$

If existed $$y_n \in A$$ such that $$y_n \to x$$ then $$y_n$$ is either the whole sequence $$x_n$$ or a subsequence of $$x_n$$ and converges to $$y$$ and to $$x$$

So from the uniqueness of limit in a metric space $$y=x$$ which is a contradiction.

Thus the set of all the limit points of $$A$$ is $$A \cup \{x\}$$

• Your arguments just show that $A \cup \{x \} \subseteq \overline A$ and that is the easy part and you don't even need to employ the characterization by sequences to see that. – Adriano Oct 2 '19 at 9:09
• @Adriano My argument shows that all limit points of A is the union of A with the singleton {x}...i will edit my answer. – Marios Gretsas Oct 2 '19 at 9:11

Let $$y \in \overline{A}\setminus A$$. Then, as we are in a metric space, there is a sequence $$s$$ of the set $$A$$ whose limit is $$y$$. Observe that $$s$$ is eventually not constant, otherwise $$y\in A$$. Then w.l.o.g. we can assume that $$s$$ is a permutation of a subsequence $$s'$$ of the sequence $$A$$. As the sequence $$A$$ converges to $$x$$, the subsequence $$s'$$ converges to $$x$$ as well (this is true in every topological space). Moreover, in a metric space, every permutation of a converging sequence is a converging sequence with the same limit. Hence $$s$$ converges to $$x$$ as well and in particular $$y=x$$.