TITLE: Determine if the set is compact
QUESTION [5 upvotes]: $$S=\{1,1/2, 2/3,3/4,....\}$$
I think this is compact as it has one sequence that covers all elements in set except $1$. This sequence is $a_n=\frac n{n+1}$. This sequence converges to $1$ hence all subsequences in $S$ converge to $1$, which is in $S$. Also this is bounded in $[1/2, 1]$.

REPLY [4 votes]: Directly from the definition of compactness:
Let be $(V_i)_{i\in I}$ an open cover of $S$. Let be $i_0$ s.t. $1\in V_{i_0}$.
As $n/(n+1)\to 1$, $\exists n_0\in\Bbb N$ s.t. $n\ge n_0\implies n/(n+1)\in V_{i_0}$.
Finally, for each $n < n_0$ take $i_n$ s. t. $n/(n+1)\in V_{i_n}$.
The family $\{V_{i_0}, V_{i_1},\dots,V_{i_{n_0-1}}\}$ is a finite open subcover of $S$.