Prove that $\lbrace x_1, x_2, x_3, . . . \rbrace \cup \lbrace x \rbrace$ is sequentially compact.

Question

Let $(x_n)$ be a sequence in a metric space converging to a point $x$. Prove that $S=\lbrace x_1, x_2, x_3, . . . \rbrace \cup \lbrace x \rbrace$ is sequentially compact.

I understand that there already exists a similar question (Show that $\lbrace x_n : n \in \mathbb{N} \rbrace \cup \lbrace x \rbrace$ is a compact subset of $(X,d)$). But In that question, the answers are all using the notion of open cover, which I have not learnt yet. So in my question, I need to prove that every sequence from $S$ has a subsequence which converges to an element of $S$. I am not sure how to do that. Can someone help me (with a few details), please?

Answer 1

Hint: For convenience, define $x_{\infty} := x$. Then you have a 'rank' map $r : S \to \mathbb{N} \cup \{\infty \} : x_k \mapsto k$. Now, let $y : \mathbb{N} \to S : k \mapsto y_k$ be a sequence. You can consider the compose map $r \circ y : \mathbb{N} \to \mathbb{N} \cup \{\infty\}$, which allows you somewhat to count how many times the sequence 'passes by' any point of $S$.

Here are two (non mutually exclusive) possibilities (you have to prove this): Either the map $r \circ y$ is constant on a subset $X \subset \mathbb{N}$ of infinite cardinality or it is strictly increasing on a subset $Y \subset \mathbb{N}$ (of infinite cardinality). To prove this, it might be useful to prove that you can always find a subset $Z \subset \mathbb{N}$ of infinite cardinality on which the function $y$ is a (not necessarily strictly) increasing function; This observation underlies the second part of timon's comment to your question$^1$. In the first case, the subsequence $ \left. y \right|_X$ is constant (and thus converges); In the second case, one has to show that the subsequence $\left. y \right|_Y$ converges to $x_{\infty}$ (which relies on the first part of timon's comment).

$^1$ To be a bit picky about timon's comment, I point out for instance that the sequence $(y_k = x_{\infty})_k$ is a sequence in $S$, but not a subsequence of the sequence $(x_n)_n$. The composition $x \circ r \circ y$ formalises in what sense $y$ could be (generally abusively) understood as a subsequence of $x$.

Answer 2

We will prove that the set $M$ is sequence compact. To do this, we need to show that every sequence from $M$ has a convergent subsequence. To this end, let $(y_n)_n \subseteq M$ be any sequence. We distinguish the following two cases:

(a) The sequence takes a value in $M$ any number of times. Thus the sequence $(y_n)_n$ automatically has a constant subsequence which is obviously convergent.

(b) The sequence takes any value from $M$ only finitely often. Let us define for $k \in \mathbb{N}$ the set $$ M_k=\left\{x_j \in X \mid j \in\{1, \ldots, k\}\right\}, $$ then for each $k \in \mathbb{N}$ there exists at least one index $$ N_k \in \mathbb{N}: x_n \notin M_{N_k}\quad \forall\, n \geq N_k. $$ Now let $\varepsilon>0$ be arbitrary. Since the sequence $ (y_n )_n$ converges to $a$ by precondition, $$ \exists\, K \in \mathbb{N}: d(y_n-a)<\varepsilon\quad \forall\,n \geq K. $$ Thus, for $n \geq N_K$ it follows. $$ d(y_n-a)=d(x_n-a)<\varepsilon, $$ that is, the subsequence $(y_n)_n$ also converges to $a$.

Thus, as desired, we have shown that the set $M$ is compact.