Show $K$ is Compact.

Question

I'm new to Real Analysis and didn't know how to start this question:

Let $x_n \rightarrow x$ in $(M, d)$. Let $K = \{x\} \cup \{x_n : n\in\mathbb{N}\}$. Show that $K$ is compact.

I wanted to try doing this by showing that for every open cover, there is a finite subcover.

So let $\{U_\alpha : \alpha \in I\}$ be an open cover for $K$... So one of the open sets of the subcover contains $x$, so assume that $U_0$ contains $x$.

Now I'm just stuck here, I tried doing multiple things but... I don't get anywhere with it.

Answer 1

If $x \in U_0$ then, since $U_0$ is open, there exists $\varepsilon > 0$ such that if $d(x,y) < \varepsilon$ then $y \in U_0$.

Therefore, by convergence of $(x_n)_{n \in \mathbb{N}}$, there exists $N \in \mathbb{N}$ such that if $n \ge N$ then $d(x,x_n) < \varepsilon$, and hence $x_n \in U_0$ for all $n \ge N$.

How might you obtain a finite subcover using this information?

Answer 2

Hint: $x_n\to x$ iff every neighborhood of $x$ contains all but finitely many $x_n$

This is sometimes given as an alternate definition of convergence. See if you can deduce that this is the case from the $\epsilon$-$\delta$ definition.

Answer 3

Hint If an open set $V$ contains $x$ then there is a natural number $N$ such that $a_n\in V$ for every $n>N$.

Answer 4

By the definition of convergence, you also know that $U_0$ has to contain all but finitely many of the $x_n$'s. Just pick out the finitely many open sets in your cover that contain the finitely many points not in $U_0$

Answer 5

$U_0$ is open, there exist an epsilon ball that contains infinitely many points. When producing a finite sub-cover, keep $U_0$ and for the remaining finitely many points, each corresponds to an open set that are kept in the finite sub-cover.