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I'm having some trouble with the following questions:

Let $S$ be any set and $\epsilon$ > 0. Define $T$ = {$t$ $\in$ $\mathbb{R}$ : |$t - s$| < $\epsilon$ for some $s \in S$}. Prove that T is open.

Now, again let $S$ be any set and define $V$ = {$t$ $\in$ $\mathbb{R}$ : |$t - s$| $\le$ 1 for some $s \in S$}. Is $V$ necessarily closed?

Thanks you for any help you can provide!

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Hint for 1): $T$ is the union of the open balls $B_s=\{ x :|x-s|<\epsilon\}$, $s\in S$.

Hint for 2): Consider $S=(0,1)$ in $\Bbb R$. Is $2\in V$? Is $(0,2)\subseteq V$?

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1) Remember the definition of an open set: a set is open iff for any point $t$ in the set, there exists an $\epsilon > 0$ such that a ball of radius $\epsilon$ around $t$ lies completely within the set. Choose an arbitrary point $t$ within the set $S$ and see if you can find such a ball.

2) Remember that a set is closed if and only if its complement is open. In this case, its complement is $|t - s| > 1$. If you can prove the set in question #1 is open, question #2 can be proven in much the same way.

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