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I know that a subset $M$ of a metric space $(X,d)$ is open if it contains a ball about each of it points, and closed it its complement is open.

But how would I show that the set $\cap_{k\in \mathbb{N}}[-\frac{1}{k},k+1]$ in $(\mathbb{R},|.|)$ is closed and not open?

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    The set is $[0,2]$, a closed interval. Closed intervals are closed.2012-05-22

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Let $C = \cap_{k\in \mathbb{N}}[-\frac{1}{k},k+1]$, and suppose $x \notin C$. Then $\exists k$, such that $x \notin [-\frac{1}{k},k+1]$. Choose $\epsilon = \frac{1}{2} \min \{ |x+\frac{1}{k}|, |x-(k+1)|\}$. Then $B(x,\epsilon) \cap C = \emptyset$, since $C \subset [-\frac{1}{k},k+1]$.

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    Ok, tha$n$ks for your help!2012-05-21