Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

2 Answers 2

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 $(0,2)\subseteq V$?

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.