Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I can prove that $A$ is closed bounded, could any one tell me $A$ is connected and dense too?thank you.

enter image description here

$A$ is the closure in $\mathcal C[0,1]$ of the set $B$ where $$B=\{f\in\mathcal C^1[0,1]; |f(x)|\le1\text{ and }|f'(x)|\le1\text{ for all }x\in[0,1]\}.$$ Answer: closed, compact, connected, dense

share|cite|improve this question
Are you sure that it is $B$ that should be proven to have the mentioned properties, and not $A$? – Jas Ter Oct 26 '12 at 6:48
@SimenK. edited, thank you – Un Chien Andalou Oct 26 '12 at 7:39
up vote 0 down vote accepted

Compactness: Arzela-Ascoli.

Connectedness: the closure of a connected set is connected (note that $B$ is convex, hence so is $A$).

Since you didn't say dense where this is impossible to answer.

share|cite|improve this answer
Well, since $A$ is compact, it is quite hard for it to be dense anywhere but itself :-) – commenter Oct 26 '12 at 9:21

Your Answer


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.