Let $(X,d)$ be a metric space, and $(x_n)$ a sequence in $X$ such that $x_n\longrightarrow x$ is convergent. Then, the set $Y=\{x_1,x_2,…\}\cup\{x\}$ is compact in $X$.
Proof attempt: Let $\left\{A_\alpha\right\}_{\alpha\in I}$ be an open cover for $Y$. Since $x\in Y$, then we can choose an $\alpha_0\in I$ such that $x\in A_{\alpha_0}$, since $A_{\alpha_0}$ is an open set, hence there is an $\varepsilon>0$ such that the open ball $B(x,\varepsilon)\subseteq A_{\alpha_0}$. By the definition of convergence, there is an $n_0\in\mathbb{Z}_+$ such that whenever $n\ge n_0$: $d(x_n,x)<\varepsilon$, i.e., for all $n>n_0, x_n\in B(x,\varepsilon)\subseteq \bigcup_{\alpha\in I}A_\alpha$.
Now, note that for any $x_m$ with $m\in\{1,2,…,n_0\}$ there is an $\alpha_m\in I$ such that $x_m\in A_{\alpha_m}$, therefore $\{x_1,x_2,…,x_{n_0}\}\subseteq \bigcup_{i=1}^{n_0}A_{\alpha_i}$. Hence, we have that:
$$Y=\{x_1,…,x_{n_0}\}\cup\{x_m:m>n_0\}\cup\{x\}\subseteq \left(\bigcup_{i=1}^{n_0}A_{\alpha_i}\right)\cup B(x,\varepsilon)\subseteq\bigcup_{\alpha\in I}A_\alpha$$.
Hence $\left\{A_{\alpha_1},A_{\alpha_2},…,A_{\alpha_{n_0}},B(x,\varepsilon)\right\}$ is a finite subcollection of $\left\{A_\alpha\right\}_{\alpha\in I}$ that also covers $Y$, therefore $Y$ is compact.
Is this proof corect? If you have any comment on the proof I'd be glad to hear them.
Thanks!
Best Answer
The general thrust is good: pick an open set $A_{\alpha_0}$ containing the limit $x$, and hence an infinite tail of the sequence. Add an open set for each of the remaining sequence points, i.e. the ones that are not clearly contained in $A_{\alpha_0}$. That said, there are places in the proof where I feel the presentation could be improved.
More to the point, $x_n \in B(x, \varepsilon) \subseteq A_{\alpha_0}$. We want the tail of the sequence to belong to a specific open set in this cover, not the union of these open sets (which, by definition, contain every sequence point and the limit).
This is a similar issue to what's above: containment in the full union is not really the point. Instead, try this:
\begin{align} Y &= \{x_1,...,x_{n_0}\}\cup\{x_m:m>n_0\}\cup\{x\} \\ &\subseteq \left(\bigcup_{i=1}^{n_0}A_{\alpha_i}\right)\cup B(x,\varepsilon) \\ &\subseteq \left(\bigcup_{i=1}^{n_0}A_{\alpha_i}\right)\cup A_{\alpha_0} \\ &= \bigcup_{i=0}^{n_0}A_{\alpha_i}. \end{align}
If the idea is to point out that each $A_{\alpha_i}$ belongs to the open cover, then I would first point out that saying $B \subseteq \bigcup_{\alpha\in I}A_\alpha$ is a weaker statement than saying $B = A_\alpha$ for some $\alpha \in I$. I also think that the fact that this holds for $B = A_{\alpha_i}$ is clear enough that you don't need to explicitly point it out in the proof.
This set should be $$\left\{A_{\alpha_0}, A_{\alpha_1},A_{\alpha_2},...,A_{\alpha_{n_0}}\right\}$$ instead. There's no reason to believe that $B(x; \varepsilon)$ is one of the open sets in the cover. By construction, it is contained in $A_{\alpha_0}$, but it's not necessarily the same set, nor is it necessarily one of the other sets in $\{A_\alpha\}_{\alpha \in I}$. So, putting it in a subcollection may not technically be correct.
Other than these small corrections, the proof is good!