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

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$A$ 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

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Are you sure that it is $B$ that should be proven to have the mentioned properties, and not $A$? –  Simen K. Oct 26 '12 at 6:48
    
@SimenK. edited, thank you –  Bunuelian Trick Oct 26 '12 at 7:39
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1 Answer

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.

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