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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?

Thanks.

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    Which function f do you have in mind here on the set R?2011-08-20

1 Answers 1

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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.2011-08-20