Is the Closure of a Totally Bounded Set Also Totally Bounded?

[sazanda]

Homework Statement

Let S be a subset of C. Prove that S is precompact if and only
if S(closure) is compact.

Homework Equations

I have already showed if S(closure) compact, then S is precompact
how can I show if S is precompact, then S(closure) is compact?

The Attempt at a Solution

 

[Bacle]

What is your definition ofS being precompact? The one I knew is that the closure of S is compact.

 

[micromass]

What is your definition ofS being precompact? The one I knew is that the closure of S is compact.

I think the OP means totally bounded. I've seen this terminology used before...

Anyway, you are working in \mathbb{C}. So try to prove that the closure of a totally bounded set is (totally) bounded. Then use Heine-Borel.

 

[sazanda]

Definition:
A set S is precompact if every ε>0 then S can be covered by finitely many discs of radius ε .

 

[sazanda]

I think the OP means totally bounded. I've seen this terminology used before...

Anyway, you are working in \mathbb{C}. So try to prove that the closure of a totally bounded set is (totally) bounded. Then use Heine-Borel.

We didn't cover totally boundedness. I think we should use definition of precompactess.

 

[Bacle]

Sazanda:
The definition you gave is the same as that of totally-bounded. I mean, totally.

 

[sazanda]

totally bounded

I think the OP means totally bounded. I've seen this terminology used before...

Anyway, you are working in \mathbb{C}. So try to prove that the closure of a totally bounded set is (totally) bounded. Then use Heine-Borel.

how Can I show that the closure of a totally bounded set is (totally) bounded?

solution Tried:

Assume S is totally bounded. then for very ε>0 there are finitely many discs (O=Union of finitely many discs) that covers S let x be a limit points of S that is in S closure. but not in S.
hence x is in O (how can I show this?)
So x is in O for all x in S closure.
Hence S closure is totally bounded.

Am I on the right track?