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

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SUMMARY

The discussion centers on proving that the closure of a totally bounded set in the complex numbers, denoted as S(closure), is compact if and only if S is precompact. The participants clarify that precompactness is equivalent to total boundedness, emphasizing the use of the Heine-Borel theorem. The proof strategy involves demonstrating that for every ε>0, S can be covered by finitely many discs of radius ε, which leads to the conclusion that S(closure) is also totally bounded.

PREREQUISITES
  • Understanding of precompactness and total boundedness in metric spaces.
  • Familiarity with the Heine-Borel theorem in the context of complex numbers.
  • Knowledge of limit points and their properties in topology.
  • Basic definitions of open covers and discs in metric spaces.
NEXT STEPS
  • Study the Heine-Borel theorem and its implications for compactness in metric spaces.
  • Explore the definitions and properties of totally bounded sets in depth.
  • Learn about limit points and their role in the closure of sets.
  • Investigate examples of precompact sets in \mathbb{C} to solidify understanding.
USEFUL FOR

Mathematics students, particularly those studying topology and analysis, as well as educators looking to clarify concepts of compactness and boundedness in metric spaces.

sazanda
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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

 
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What is your definition ofS being precompact? The one I knew is that the closure of S is compact.
 
Bacle said:
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.
 
Definition:
A set S is precompact if every ε>0 then S can be covered by finitely many discs of radius ε .
 
micromass said:
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.
 
Sazanda:
The definition you gave is the same as that of totally-bounded. I mean, totally.
 
totally bounded

micromass said:
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?
 

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