Compact by Sequences

Question

Let $(M,d)$ be a Metric Space and Let $X:\mathbb N \to M$ be a converegent sequence to $l \in M$

Prove that $H = \{X_n\}^\infty_{n = 1}\cup \{l\}$ is compact by sequences

I know that to prove that a set is compact by sequences, then we need that every sequences admits a convergente subsequence, and I know that the Heine Borel theorem is not applicable for all metric spaces. How should I proceed then?

Answer 1

Let $(h_n)_{n\in\mathbb N}$ be a sequence of elements of $H$. Then:

• if $h_n=l$ for infinitely many natural numbers $n$, then you have a subsequence that converges to $l$;
• if for some $m\in\mathbb N$, you have infinitely many natural numbers $n$ such that $h_n=X_m$, then you have a subsequence that converges to $X_m$;
• otherwise, there is some $h_{n_1}$ which is equal to some $X_{m_1}$ with $m_1\geqslant1$, there is some $h_{n_2}$ which is equal to some $X_{m_2}$ with $m_2\geqslant2$, and so on. Therefore, $\lim_{k\to\infty}h_{n_k}=l$.

Answer 2

All but one of the points of the space $H$ are isolated points, i.e. you can put an open ball of positive radius about each of them, that contains none of the others.

So none of those points can be a limit of a subsequence except by occurring infinitely many times in the sequence. If none of them occurs infinitely many times, then for every finite set $S$ of isolated points, there is some index $N$ such that for all $n\ge N,$ the $n$th term of the sequence is not in $S$. Now look at an open ball centered at the point $\ell.$ No matter how small its radius is, it must include all but finitely many points of the space. From this and the foregoing remarks you can deduce that a sequence with no subsequence converging to any of the isolated points must converge to $\ell.$

Answer 3

Let $(y_n)_{n \in \mathbb N}$ be a sequence in $H$. If $y_n = l$ for infinitely many $n \in \mathbb N$, you can simply choose a subsequence which converges to $l \in H$. So assume that $y_n = l$ only for finitely many $n \in \mathbb N$. Then for each $\epsilon > 0$, there are infinitely many $y_n \in \mathbb B(l, \epsilon)$. Choose $y_{n_1} \in \mathbb B(l, 1)$. Now choose $y_{n_2} \in \mathbb B\left((l, \frac 1 2\right)$, such that $n_2 > n_1$. By continuing this process you get a subsequence $(y_{n_k})_{k \in \mathbb N}$, that converges to $l$.