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The set is in $R^1$ and consists of $0$ and the numbers $1/n$. Call it $E$.

Take a set of $n$ intervals of radius $r$, centered less than $2r$ apart and such that $\sum_{i=1}^n r \ge 1/2$. Call the union of such intervals $B$.

Suppose $G$ is an arbitrary open cover of $E$ and $A = B \cap G$. Then $A \subseteq $G and also covers $E$ with finitely many balls. Since $G$ was any open cover, the set $E$ is compact.

Is this correct and rigorous enough? The proof provided as solution says that $0$ is contained in an open set, and thus for $n$ big enough all points $1/n$ and after will be inside this open set, leaving only a finite number of points outside. The union of this set and those points is an open subcover. I feel like my proof is close enough that it should be correct but I'm not sure.

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    $\begingroup$ The set you seem to be talking about is not $[0,\frac{1}{n})$ but instead $\{\frac{1}{n}:n\in\Bbb{N}\}\cup\{0\}$. Maybe you should also write what is your initial open cover that $G$ is a subcover of and why this would give you the freedom to make the claimed choice of open intervals in the second paragraph of your post. $\endgroup$ – T. Eskin May 3 '16 at 1:07
  • $\begingroup$ Changed it, thanks. $\endgroup$ – badmax May 3 '16 at 1:10
  • $\begingroup$ Must be a typo: you start off "suppose $G$ is an open subcover..." And later you conclude by the reason "since $G$ was any open cover." surely you meant to start with an arbitrary open cover? Tbh i dont understand your proof. The sketch offered in the solution is the direct approach that uses the open cover definition and could be worth fleshing out. There's also heine-borel... $\endgroup$ – Nap D. Lover May 3 '16 at 1:21
  • $\begingroup$ Yes I meant to start with an arbitrary cover $G$. My proof is basically to cover the set $E$ with a collection of intervals of radius $r$, then take their intersection with $G$ as a finite subcover. $\endgroup$ – badmax May 3 '16 at 1:24
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    $\begingroup$ Let G = {R}. That is an open cover with one element set. A= G intersect E = the empty set as none of the open balls are R. The empty set is NOT a cover. $\endgroup$ – fleablood May 3 '16 at 1:36
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Proof. Let $G$ be an open cover of $E$.

$G$ contains an open set $G_0$ containing an $\epsilon$ neighborhood of $0$. For each $n$ we also have an open set $G_n \in G$ with $1/n \in G_n$. Let $K$ be large enough that $1/K < \epsilon$. Then $$E \subset G_0 \cup G_1 \cup G_2 \cup \cdots \cup G_K$$ Done.

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By the theorem of Heine-Borel, it suffices to show that the set E is bounded and closed. E is obviously bounded, since each element in it has absolute value $\leq 1$. Now it is also easily seen that the complement $\mathbb{R}-E$ is open: for each point in the complement we can take an intervall of sufficiently small radius around the point, so that none of the points in E are in our intervall. Hence E is also closed and we are done.

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To prove the statement directly by definition, let $\{O_\alpha\}_{α\in E}$ be an arbitrary open cover of $K$. Then for each $i\in \mathbb{N}$, \begin{equation} \frac{1}{i}\in O_{\alpha_i} ~\text{for some}~α_i\in E, \end{equation} and $0\in O_{\alpha_0}$ for some ${\alpha_0}\in E$. Since $O_{\alpha_0}$ is open, $0$ is an interior point of $O_{\alpha_0}$. Then there exists an $\epsilon>0$ such that $N_ϵ(0)\subset O_{\alpha_0}$ and there exists a positive integer $n_0$ such that $n_0 ϵ>1$. So, $1/n_0 \in O_{α_0}$ and for all $n≥n_0$, $1/n\in O_{α_0}$. Therefore, $O_{α_0} \cup O_{α_1}\cup\ldots \cup O_{α_{n_0-1}}$ is a finite subcover of $K$.

To see this—let's take $x \in K$. If $x=0$ or $x=1/j$ for some $j\in \mathbb{N}$ with $j≥n_0$, then $x\in O_{α_0}$. But if $j<n_0$, then $x∈O_{α_i}$ for some of $i=1,~2,…,~n_0-1$. Therefore, \begin{equation} K\subset \cup_{i=0}^{n_0-1} O_{α_i}. \end{equation}

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