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

  • Thread starter sazanda
  • Start date
In summary: Trying to show that the closure of a totally bounded set is (totally) bounded. I'm assuming that this is equivalent to proving that the set is compact.In summary, if a set is precompact then its closure is also precompact.
  • #1
sazanda
12
0

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

 
Physics news on Phys.org
  • #2
What is your definition ofS being precompact? The one I knew is that the closure of S is compact.
 
  • #3
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 [itex]\mathbb{C}[/itex]. So try to prove that the closure of a totally bounded set is (totally) bounded. Then use Heine-Borel.
 
  • #4
Definition:
A set S is precompact if every ε>0 then S can be covered by finitely many discs of radius ε .
 
  • #5
micromass said:
I think the OP means totally bounded. I've seen this terminology used before...

Anyway, you are working in [itex]\mathbb{C}[/itex]. 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.
 
  • #6
Sazanda:
The definition you gave is the same as that of totally-bounded. I mean, totally.
 
  • #7
totally bounded

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

Anyway, you are working in [itex]\mathbb{C}[/itex]. 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?
 

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

What is the definition of compactness?

Compactness is a mathematical property of a set that indicates how much the set "fits" inside a larger space. In simpler terms, it measures how tightly the points in a set are clustered together.

What does "iff" mean in the phrase "Compactness iff precompact"?

"Iff" is a mathematical shorthand for "if and only if". In this context, it means that compactness and precompactness are equivalent, or they are both true at the same time.

Can you explain precompactness?

Precompactness is a weaker version of compactness that is often used in metric spaces. It means that the set can be covered by a finite number of smaller sets, each of which has a finite diameter.

How does compactness relate to limits?

Compactness is closely related to the concept of limits. In a compact set, any sequence of points will have a limit point, or a point that the sequence "approaches" as the number of points increases. In a non-compact set, this may not be the case.

What are some examples of compact and non-compact sets?

A few examples of compact sets are a closed interval on the real number line, a sphere in three dimensions, and a closed and bounded subset of a metric space. Non-compact sets include an open interval on the real number line, an infinite plane in two dimensions, and an unbounded subset of a metric space.

Similar threads

Back
Top