$$S=\{1,1/2, 2/3,3/4,....\}$$ 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 $[1/2, 1]$.
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Your reasoning is right: the space is sequentially compact, and hence compact. – Patrick Stevens Mar 27 '18 at 11:19
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Yes, this set is compact by Heine-Borel. It is closed (contains its limit point) and it is bounded.
Siong Thye Goh
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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 $S$.
Martín-Blas Pérez Pinilla
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HINT:
$1$. A set $A\subseteq \mathbb{R}$ is said to be compact iff it is closed and bounded .
$2$. If $D(A)\subseteq A$ , then $A$ is closed.
NOTE: For given set $S$ , it's all terms lie between $\frac{1}{2}$ and $1$.
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