# Using Open Covers to prove a set is compact

If $$(x_n)$$ is a sequence converging to a limit $$L$$ , let $$Y = \lbrace x_n:n \epsilon \mathbb{N}\rbrace \cup\lbrace L \rbrace$$. Using open covers, prove that Y is compact.

Any help would be appreciated, have no idea where to start.

• "using open covers" tells you that you have to start with "Let $U$ be an open cover of $Y$. Then..." . Can you continue from here? Do you know the definition of compactness by means of open covers? – Crostul Dec 17 '19 at 13:27
• Yes, if I can construct a finite subcover of Y from U, then Y is compact – carbonv2 Dec 17 '19 at 13:29
• Hint: At least one of your covering sets contains $L$. – quasi Dec 17 '19 at 13:30
• Using the fact that $x_n$ converges to $L$, then whenever you have an open neighbourhood of $L$, only finitely many elements of $x_n$ don't belong to it. This allows you to select only finitely many open sets from the open cover. – Crostul Dec 17 '19 at 13:33
• So then, we would select N number of open sets from the open cover and take the union of them alongside the union of the epsilon neighborhood of L to construct our finite subcover? – carbonv2 Dec 17 '19 at 13:42

Let $$\{O_i: i \in I\}$$ be any open cover of $$Y$$.

This means there is some $$i_0 \in I$$ such that $$L \in O_{i_0}$$, as $$L \in Y$$ must be covered.

By definition of convergence and as $$O_{i_0}$$ is an open neighbourhood of the limit $$L$$, there is some $$N \in \Bbb N$$ such that

$$\forall n \ge N: x_n \in O_{i_0}$$

So that one open set already contains almost all points of $$Y$$, and for each $$n < N$$ we can find $$i(n) \in I$$ such that $$x_n \in O_{i(n)}$$ (all points must be covered!) and hence we have covered all points of $$Y$$ by the finite subcover

$$\{O_{i_0}\} \cup \{O_{i(n)}: n < N\}$$

and as the starting cover was arbitrary, $$Y$$ is compact.

• Elegant answer with topology. Thank you very much. – carbonv2 Dec 17 '19 at 20:41

Let $$\Lambda:=\{A_i\}_i$$ be an open covering of $$(x_n)_n\cup \{L\}\subseteq X$$.

Then $$L\in A_j$$ for some $$j$$ and this means there exists $$N\in\mathbb{N}$$ such that $$x_n\in A_j$$ for each $$n\geq N$$.

Moreover $$x_m\in A_{i(m)}$$ for some $$i(m)$$, $$m\leq N$$, and this permit us to say that

$$(x_n)_n\cup \{L\}\subseteq A_{i(1)}\cup \cdots \cup A_{i(N)}\cup A_j$$