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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?

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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 $y=x$.

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  • $\begingroup$ 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. $\endgroup$ – Adriano Oct 2 '19 at 9:11
  • $\begingroup$ @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 $n_1<n_2<...$. $\endgroup$ – Kavi Rama Murthy Oct 2 '19 at 9:16
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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 $A \cup \{x\}$

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  • $\begingroup$ 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. $\endgroup$ – Adriano Oct 2 '19 at 9:09
  • $\begingroup$ @Adriano My argument shows that all limit points of A is the union of A with the singleton {x}...i will edit my answer. $\endgroup$ – Marios Gretsas Oct 2 '19 at 9:11
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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 $y=x$.

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