Let $A$ be a subset of $R$ which consist of $0$ and the numbers $\frac{1}{n}$, for $n=1,2,3,\dots$. I want to prove that $K$ is compact directly from the definition of compact.

So, given any open cover of $A$, I should be able to find a finite subcover. Proving a set is compact is much difficult than proving not compact. I have find a process of finding a finite sub cover for every open cover which means I need to find some common property of every open cover.

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    $\begingroup$ Hint: every cover contains a neighbourhood of $0$. What can you say about how much of $A$ this neighbourhood covers? $\endgroup$ Sep 6, 2013 at 9:43

1 Answer 1


Let $\mathcal U$ be a cover of $A$. As $\mathcal{U}$ is a cover, there exists an open $U\in\mathcal{U}$ such that $0\in U$. As $U$ is open, there exists an $\epsilon>0$ such that for all $r\in[0,\epsilon)$, $r\in U$. Now, the set $A$ only has finitely many elements greater than or equal to $\epsilon$. Can you see why this means we only need to choose a finite number of other open sets in $\mathcal{U}$ to cover the rest of $A$?


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