Totally bounded subset in a metric space

In summary, the author proves that a metric space is totally bounded if and only if it has a finite but dense subset. This covering exists no matter the choice of the radius of the subset.
  • #1
tarheelborn
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Homework Statement



Suppose [tex]M[/tex] is a metric space and [tex]A \subseteq M[/tex]. Then [tex]A[/tex] is totally bounded if and only if, for every [tex]\epsilon >0[/tex], there is a finite [tex]\epsilon[/tex]-dense subset of [tex]A[/tex].

Homework Equations





The Attempt at a Solution



I have already done the [tex]\Rightarrow[/tex] but need to verify the other half:
[tex](\Leftarrow )[/tex]: Now suppose that for [tex]\epsilon > 0[/tex], [tex]A[/tex] has a finite [tex]\epsilon[/tex]-dense subset. I must prove that [tex]A[/tex] is totally bounded. Since there is an [tex]\epsilon[/tex]-dense set in [tex]A[/tex], say [tex] \{ x_1, x_2, \cdots, x_n \} [/tex] is [tex]\epsilon[/tex]-dense in [tex]A[/tex], then [tex]B[x_i; \epsilon], \cdots, B[x_n; \epsilon][/tex] form a covering of [tex]A[/tex] by sets of diameter [tex]< \epsilon[/tex]. Hence [tex]A[/tex] is totally bounded.

Does this work?
 
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  • #2
This is an essentially correct argument, but for one small technical problem and a few stylistic issues.

The small technical problem is that the diameter of [tex]B(x_j; \epsilon)[/tex] is not necessarily [tex]\epsilon[/tex]; what is it? (Correcting this doesn't change the gist of the argument, but you do need to make adjustments.)

As for the stylistic issues -- Because you're being asked to prove something this simple, it might not hurt to give an explicit argument that the balls [tex]\{B(x_j; \epsilon)\}[/tex] cover all of [tex]A[/tex] -- that is, given any [tex]x\in A[/tex], why does [tex]x[/tex] lie in one of the balls [tex]B(x_j; \epsilon)[/tex]?

Also, you could be more careful about the quantifiers -- it should run something like: Fix [tex]\epsilon > 0[/tex]; we will exhibit a covering of [tex]A[/tex] by finitely many sets of diameter [tex]\leq \epsilon[/tex]. We know that there exists a finite something-dense set in [tex]A[/tex], so ... (continue from here). The important point here is that you need to show the covering exists no matter the choice of [tex]\epsilon[/tex], which isn't clear from what you wrote above.
 
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  • #3
I see. Thank you.
 

1. What is a totally bounded subset in a metric space?

A totally bounded subset in a metric space is a subset of a metric space where every finite set of points can be covered by a finite number of balls of a given size. In other words, the subset can be "totally" contained within a finite number of smaller balls.

2. How is total boundedness different from boundedness?

While boundedness refers to the property of a set to have a finite upper and lower bound, total boundedness refers to the property of a set to be able to be covered by a finite number of smaller balls. A bounded set may not necessarily be totally bounded, but a totally bounded set is always bounded.

3. Can a subset be totally bounded if the entire metric space is not?

Yes, it is possible for a subset to be totally bounded even if the entire metric space is not. This is because the total boundedness of a subset only depends on the subset itself and not on the entire metric space.

4. How is total boundedness related to compactness?

Total boundedness is a necessary condition for compactness. This means that if a subset is totally bounded, then it is also compact. However, the converse is not always true - a subset can be compact without being totally bounded.

5. What are some examples of totally bounded subsets in a metric space?

Some examples of totally bounded subsets in a metric space include a closed interval in the real numbers, a closed ball in a Euclidean space, and a finite set of points in any metric space.

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