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For example, [1,3] is a closed and bounded subset of metric space $\mathbb{R}$, so the set should be compact. But consider the open cover $(0.9,1.1)\cup(2.9, 3.1)\cup(2-\frac{1}{n}, 2+\frac{1}{n})$, where $n \in \mathbb{Z^+} $is from 1 to $\infty$, this open cover is infinite. So this set should not be compact.

What's wrong?


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I'm not sure whether this is a problem of language or not, but: Having an infinite open cover is fine. If you can't find a finite subcover then we are in trouble, but Prof Scott has exhibited one. – Dylan Moreland Aug 20 '11 at 2:35
@Moreland: Yes, it's a problem of language. I thought the open cover itself should not be infinite, but the fact is the open cover itself could be infinite, but it must have(contains) a finite subcover. – Jichao Aug 20 '11 at 2:45
Bingo,Dylan. In fact,here's an exercise for you,Jichao-The open cover you've given is generated by an infinite sequence of open subsets of (x,3] where x is some real number between 0 and 1. Can you define a subsequence of this sequence which is finite and yet still covers [1,3]? Why or why not? If the answer's yes,clearly this set is compact,yes? See if you can get the answer yourself-it's not hard. – Mathemagician1234 Aug 20 '11 at 2:53
@Mathemagician1234: Sorry, I could not understand your question well. Do you mean whether I could find a finite subcover of the open cover I post in the original question. If that, then Brian have done that. – Jichao Aug 20 '11 at 3:06
@Jichao Yes,but I was suggesting that YOU find a finite open subcover of the cover you gave. Math is not a spectator sport. – Mathemagician1234 Aug 20 '11 at 3:10
up vote 3 down vote accepted

Your cover consists of exactly one set, so it’s already finite. Perhaps you meant the cover to be $\left\{(0.9,1.1),(2.9,3.1)\right\}\cup \left\{\left(2-\frac1n,2+\frac1n \right):n \in \mathbb{Z}^+\right\}$, but there’s still no problem: this has $\left\{(0.9,1.1),(2.9,3.1),(1,3)\right\}$ as a finite subcover of $[1,3]$. Indeed, it’s clear even without reference to the specific interval $[1,3]$ that $\left(2-\frac1n,2+\frac1n \right) \subseteq (1,3)$ for all $n \in \mathbb{Z}^+$, so the intervals of this form with $n>1$ are clearly redundant.

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@Jichao And Brian gave you one possible correct answer,so you don't have to.Oh well. My point was in topology,understanding definitions is absolutely critical,more so then in most branches of mathematics. – Mathemagician1234 Aug 20 '11 at 2:55

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