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Find an open cover $\mathcal{G}$ of the set $E = \{ 1/n : n \in \mathbb{Z^{+}}\}$ such that any proper subset of $\mathcal{G}$ is not an open cover of $E$.

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What topology? The one inherited from $\mathbb R$? – Alex Becker Mar 29 '12 at 1:54
Yes we are working on the real number line. – John Lombardi Mar 29 '12 at 1:59
up vote 3 down vote accepted

Hint: for each positive integer $n$, find an open interval $O_n$ containing $1\over n$ that contains no other point of $E$. (I'm assuming you're working in $\Bbb R$ with the usual topology.)

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The easiest way is to make sure that every member of $\mathcal{G}$ is needed in order to cover $E$. That means that for each $G\in\mathcal{G}$ you want a point $e_G\in E$ that is covered by $G$ and by no other member of $\mathcal{G}$. One easy way to do this is to arrange matters so that each member of $\mathcal{G}$ contains exactly one point of $E$, a different point for each $G\in\mathcal{G}$.

What if $E$ were the set of positive integers? Can you find open sets $G_n$ for $n\in\Bbb Z^+$ such that $G_n\cap\Bbb Z^+=\{n\}$? That shouldn’t be too hard. And once you’ve seen the idea, it shouldn’t be too hard to adapt it to your actual problem.

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You can cover each point of $E$ with an interval so small that it contains no other points in $E$. Since this looks like it might be homework, I'll leave you to work out the cover explicitly.

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