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A set $M \subseteq \mathbb{R}$ is called open if $\mathbb{R} \setminus M$ is closed. How do you prove that $(2n -1/2, 2n + 1/2)$ is open?

I‘m confused because it means $2n - 1/2 < 2n + 1/2$ but how can I write it as a complement?

In this case, these are all numbers outside the interval $(2n - 1/2, 2n + 1/2)$. Is it then $[-\infty, 2n-1/2)$ and $(2n+1/2, +\infty]$?

user1729
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  • I presume $n$ is fixed here? Your complement is almost correct, but not quite - you have the square and round brackets the wrong way round ($\pm\infty\not\in\mathbb{R}$, and $2n\pm1/2$ are in neither set!). – user1729 May 29 '23 at 17:02

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$$\mathbb{R}\setminus{(2n-1/2, 2n+1/2)}=(-\infty,2n-1/2]\cup[2n+1/2,+\infty),$$ which is the union of two closed sets (in the standard topology)

Guillermo García Sáez
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  • Okay, thank you very much. But I‘m wondering if we look at the limes than we have (-∞,+ ∞]∪ [+∞, + ∞] right?, so for every convergent sequence in A there is also its limit lies in A. –  May 29 '23 at 14:15