For a given metric space, to show the set $A = \{ p, p_1, p_2, \ldots \}$ is closed if $p_n \rightarrow p$. The only property of closed sets I am working with here is:

A subset $A$ is closed iff the limit of each convergent sequence of points in $A$ is also in $A$.

So let $(q_n)_{n\geq 1} = (q_1, q_2, q_3, \ldots )$ be a convergent sequence and for each integer $n \geq 1$ require that the term $q_n \in A$.
One sees readily that $(q_n)_{n \geq 1}$ is not necessarily a subsequence of $(p_n)_{n \geq 1}$.
$(q_n)_{n\geq 1}$ could certainly be a constant sequence such as $(p_3, p_3, p_3, \ldots )$, and that is no subsequence of $(p_n)_{n \geq 1}$.
Or why not some peculiar non-subsequence such as

$ (q_n)_{n\geq 1} = (p_2,p_1,\underbrace{p_3,p_3,\ldots, p_3}_{1000 \text{ terms} }, p_6, p_5, p_8, p_7, p_{10}, p_9, p_{12}, p_{11}, \ldots)$

For the case where $(q_n)_{n \geq 1}$ is a constant sequence, it is obvious the limit is in $A$.
For the case where $(q_n)_{n \geq 1}$ is some subsequence of $(p_n)_{n\geq 1}$, the limit is $p$ and thus in $A$.
But clearly these two do not together capture all possible convergent sequence in $A$. And it's those other remaining (and peculiar) sequences I'm struggling to account for.
Any thoughts on how we can attack the remaining possible convergent sequences appreciated!

I realize there might be easier ways to show this with other or more general definitions, but I'm strictly working with the definition provided. Also, I have proven this before by showing $A^c$ is open, but I wanted to attempt to show it by means of the above definition exclusively.

 A: Let $X$ be the whole metric space and take $x\in X\setminus A$ (if no such $x$ exists, then $A=X$ and therefore $A$ is closed). Consider the ball $B_\varepsilon(x)$, where $\varepsilon=\frac12d(x,p)$. If $N$ is large enough, then$$n\geqslant N\implies d(p_n,p)<\varepsilon\implies d(p_n,x)>\varepsilon.$$So, the ball $B_\varepsilon(x)$ only contains finitely many $p_n$'s. Since $x\notin A$, this proves that no sequence of elements of $A$ converges to $x$. So, every convergent sequence of elements of $A$ converges to an element of $A$, and this proves that $A$ is closed.
A: To understand $(q_n)$ better, we need to use the fact that it converges to something.
Suppose it converges to some point $q$. There are two cases:

*

*$q_n = q$ for some $n$. In this case, since $q_n \in A$, we also have $q \in A$.

*$q_n \ne q$ for all $n$. In this case, we should be able to pick out a subsequence of $(q_n)$ that (1) still converges to $q$, and (2) is a subsequence of $(p_n)$ as well. (Try this on your own if you like, though I'm about to tell you how.) Therefore $q=p$, so $q \in A$.

To do this, define a subsequence $(r_n)$ of $(q_n)$ as follows. Let $r_1 = q_1$. For each subsequent term:

*

*Suppose $r_{n-1}$ was $p_{m}$ for some $m$.

*Let $\epsilon = \min\{\frac1n, d(p_1,q), d(p_2,q), \dots, d(p_m,q)\}$.

*Let $r_n$ be the next element of $(q_n)$ after $r_{n-1}$ such that $d(r_n,q) < \epsilon$.

This is a subsequence of $(q_n)$ by construction. It is a subsequence of $(p_n)$ because $r_n \in A$ and yet $r_n$ cannot be any of $p_1, p_2, \dots, p_m$, so $r_n$ comes after $r_{n-1}$ in $(p_n)$. Finally, it converges to $q$ because $d(r_n, q) < \frac1n$.
A: Students often fail to use the fact that in a metric space $(X,d)$ we have $p_n\to p$ iff $\{n\in \Bbb N: p_n\not \in U\}$ is finite whenever $U$ is a nbhd of $p$.
If $q\in A^c$ let $r=\frac {1}{2}d(q,p).$ Let $U=B_d(p,r).$ Let $V=\{n\in \Bbb N: p_n\not \in U\}.$ Let $W =A \setminus (U\cup \{p\}).$ Then $W$ is finite because $V$ is finite and $W=\{p_n:n\in V\}.$
Now  $q\not\in W$ [as $q\in A^c$ and $W\subset A$] and $W$ is finite so there exists $s\in (0,r]$ such that $B_d(q,s)$ is disjoint from $W.$ So no sequence of members of $A$ can converge to $q$ because $d(q,x)>s$ for any $x\in A.$
