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Is the infinite union of intervals $(\frac{1}{n}, 1]$ where $n \in \mathbb{N}$ i.e. $\bigcup\limits_{n=1}^\infty (\frac{1}{n}, 1]$ a closed subset of $\mathbb{R}$?

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Does the set contain the sequence $\{x_n\}$ where $x_n= \frac{1}{n}$? Does it contain the limit of the sequence? – Arturo Magidin Mar 4 '12 at 19:11

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This union is not closed, it is actually the interval $(0,1]$ even if you replace the sets $(\frac{1}{n},1]$ by $[\frac{1}{n},1]$.

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$\{[\frac{1}{n},1]\ |\ n\in\mathbb{N}\}$ is also a great example of an infinite union of closed sets not being closed – you Mar 4 '12 at 19:21

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