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.

  • $\begingroup$ "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? $\endgroup$ – Crostul Dec 17 '19 at 13:27
  • $\begingroup$ Yes, if I can construct a finite subcover of Y from U, then Y is compact $\endgroup$ – carbonv2 Dec 17 '19 at 13:29
  • $\begingroup$ Hint: At least one of your covering sets contains $L$. $\endgroup$ – quasi Dec 17 '19 at 13:30
  • $\begingroup$ 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. $\endgroup$ – Crostul Dec 17 '19 at 13:33
  • $\begingroup$ 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? $\endgroup$ – 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.

  • $\begingroup$ Elegant answer with topology. Thank you very much. $\endgroup$ – 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

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


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