Using the topological definition of compact, namely that every open cover admits a finite open subcover, I was hoping someone can provide me with the finite sub cover of $\{(\frac{1}{n},1-\frac{1}{n}):n\in \mathbb{N}\}$ of the interval $[0,1]$. (This $\textit{is}$ an open cover of that interval, right?). To me, it seems as though as soon as we choose a particular N to stop at, we wouldn't have covered the whole interval- yet I know the compactness of this interval is a fundamental fact of topology.
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That cover doesn't make sense because if you take $n = 1$, you get the interval $(1,0)$, which doesn't make sense. Setting that aside, the points $x = 0,1$ are in none of those intervals, so it's definitely not an open cover of $[0,1]$. – Ethan Alwaise Nov 27 '15 at 03:31
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Your proposed open cover of $[0, 1]$ doesn't include the endpoints $0$ and $1$. – Rob Arthan Nov 27 '15 at 03:31
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Which of those sets cover $1$ or $0$? – dafinguzman Nov 27 '15 at 03:31
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But doesn't $\lim\limits_{n\to\infty}\frac{1}{n}=0$? – wan greksi Nov 27 '15 at 03:57
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the union of sets only contain the elements of each set. Since 0 is not in any of the sets, it cannot be in the union by definition of union. – Alan Nov 27 '15 at 04:23
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You're right that there's no finite subcover. But that's not a problem, because it's not a cover at all! The union of all the sets $(1/n,1-1/n)$ is only $(0,1)$, not $[0,1]$. The points $0$ and $1$ are not in any of your open sets.
Eric Wofsey
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Why doesn't that imply the left side of the interval will eventually hit 0? i thought closed sets contain all limit points – wan greksi Nov 27 '15 at 04:08
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3@wangreksi For all $n$ you have $0 \notin (\frac{1}{n} , 1- \frac{1}{n})$, so that $0$ does not belong to the union of those. WHat is true is that it belongs to the closure of the union, but this condition is not what is meant by open cover of a space. – Crostul Nov 27 '15 at 08:44