Basic question on 'bounded implies totally bounded'

In summary, the concept of "bounded implies totally bounded" explores the relationship between bounded sets in metric spaces and their total boundedness. A set is bounded if it can be contained within a finite distance, while it is totally bounded if it can be covered by finitely many open balls of any given radius. The statement questions whether every bounded set in a metric space is necessarily totally bounded, highlighting the need for additional conditions, such as completeness, to establish this implication.
  • #1
psie
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TL;DR Summary
I am reading a proof of the fact that if is bounded, then it is totally bounded. This proof is from Gamelin's and Greene's book on topology. I have a question about the proof.
Recall, a set is totally bounded if for each , there exists a finite number of open balls of radius that cover .

Proof: Since is bounded, it is contained in some cube and it suffices to show that is totally bounded since a subspace of a totally bounded metric space is totally bounded. For this, let . Then the balls cover , where ranges over all integral lattice points of which satisfy , (recall, an integral lattice point is a point whose coordinates are integers). Since there are only finitely many such lattice points, is totally bounded.

Question: How can I verify that the balls cover ? In particular, why the condition , ?
 
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  • #2
psie said:
TL;DR Summary: I am reading a proof of the fact that if is bounded, then it is totally bounded. This proof is from Gamelin's and Greene's book on topology. I have a question about the proof.

Recall, a set is totally bounded if for each , there exists a finite number of open balls of radius that cover .



Question: How can I verify that the balls cover ? In particular, why the condition , ?
Why not work through an example for and .
 
  • #3
PeroK said:
Why not work through an example for and .
I have tried, but I don't see how to generalize. The condition is kind of clear; if and , it does not suffice to cover by open balls of radius centered at the points . We need more points to center at.

Here's my reasoning so far: if , then for . Let satisfy for . Now I want to show , but I am not sure how to do that. I suspect one would like to consider a third point and use the triangle inequality somehow. I am not sure.
 
  • #4
I must admit I don't see that works. I reckon you need the lattice points for each coordinate to be apart.
 
  • #5
it looks to me as if 2b should be just b (assuming b is an integer, which you can do). I.e. one can cover the cube by balls of unit radius centered at the integral lattice points within the (closed) cube. This is essentially the general case. I.e. similarly one can cover the cube by balls of radius 1/3 centered at points in the cube which are 1/3 of integer lattice points.... one can cover the cube by balls of radius e, centered at points within the cube which are e times an integer lattice point.

in post #3, you need also points centered at ±1/3, ±2/3,....,±15/3. i.e. all j such that (1/3)j ≤ 5.

Based on this example, I would suggest choosing another book.
 
  • #6
mathwonk said:
it looks to me as if 2b should be just b (assuming b is an integer, which you can do). I.e. one can cover the cube by balls of unit radius centered at the integral lattice points within the (closed) cube. This is essentially the general case. I.e. similarly one can cover the cube by balls of radius 1/3 centered at points in the cube which are 1/3 of integer lattice points.... one can cover the cube by balls of radius e, centered at points within the cube which are e times an integer lattice point.

in post #3, you need also points centered at ±1/3, ±2/3,....,±15/3. i.e. all j such that (1/3)j ≤ 5.

Based on this example, I would suggest choosing another book.
Doesn't that factor of 3 increase with the dimension ?
 
  • #7
I don't think so. in dim 2, you need balls centered at (0,0), (±1/3,±1/3), (±1/3,±2/3), (±2/3,±1/3), (±2/3,±2/3), ....

all you are doing is marking off new lattice points in the unit cube, at every point of form (j/m,.....,k/m). so when e = 1/m, you would have I guess, 1 + m^n such points in the cube [0,1]^n.

He also makes it more confusing by using negative numbers. One can translate any bounded set into an expansion of [0,1]^n.

Basically he is making an easy argument hard. Just note that unit radius balls, centered at integer lattice points, cover [0,C]^n, for any integer C, and then just scale that down by multiplying by e, where C = b/e.

oops, now I see why maybe he used 2b. If e is not of form 1/m, we want to be sure our scaled down lattice points do cover the cube [0,b]^n.

maybe I'm wrong, but I would just blow up the cube [0,b]^n, by dividing by 1/e, i.e. to [0,b/e]^n. Then choose an integer m ≥ b/e, and cover the cube [0,m]^n, hence also the blown up cube [0,b/e]^n, by unit radius balls centered at integer lattice points. Then shrink everything down again by multiplying by e, to get the original cube [0,b]^n covered by balls of radius e, centered at points with coordinates of form ej, where j is an integer.

(of course it suffices to assume e = 1/k, for some integer k, so we may in fact assume b/e is itself an integer.)

does that seem ok?
 
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  • #8
mathwonk said:
I don't think so. in dim 2, you need balls centered at (0,0), (±1/3,±1/3), (±1/3,±2/3), (±2/3,±1/3), (±2/3,±2/3), ....
I was looking at a simple approach, where for every we can find a point in the lattice, such that:
Where is the linear lattice spacing. Then:
In fact, it seems that for any lattice, we can always find where we have equality. That suggests to me that we need the linear spacing to depend on .
 
  • #9
perhaps I am ignoring the fact that the diagonal of a cube goes up with n?

I think you are right. so there are two steps perhaps. one, show that for every e>0, there is some k such that the cube of radius 1/k is inscribed in a ball of radius e. two, then show that every cube of edge length b can be subdivided into small cubes of edge length 1/k.

these are both easy, but apparently also easy to get wrong, at least for me. I just eyeballed it for n=2, where 1+1 > sqrt(n).

thank you!

(or just finesse step one by using the "max norm", so that ball becomes a cube.)
 
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  • #10
so I think I finally see it, and of course it is what you said all along.

1). it suffices to consider just the unit cube [0,1]^n, since every cube can be covered by finitely many copies of that one.

2). for every integer k, we can subdivide the unit cube into finitely many (k^n) subcubes of edge length 1/k.

3). Since the sub cube of edge length 1/k has diagonal sqrt(n)/k, we can cover that subcube by one open euclidean ball of radius sqrt(n)/k, centered at the center of the sub cube, so as long as we choose k large enough so that sqrt(n)/k < e, i.e. so that k > sqrt(n)/e, we are done.

i.e. as you said in post #4, we want edge length 1/k < e/sqrt(n).
 
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  • #11
mathwonk said:
so I think I finally see it, and of course it is what you said all along.

1). it suffices to consider just the unit cube [0,1]^n, since every cube can be covered by finitely many copies of that one.

2). for every integer k, we can subdivide the unit cube into finitely many (k^n) subcubes of edge length 1/k.

3). Since the sub cube of edge length 1/k has diagonal sqrt(n)/k, we can cover that subcube by one open euclidean ball of radius sqrt(n)/k, centered at the center of the sub cube, so as long as we choose k large enough so that sqrt(n)/k < e, i.e. so that k > sqrt(n)/e, we are done.

i.e. as you said in post #4, we want edge length 1/k < e/sqrt(n).
I appreciate your solution, but how would one amend the intended proof?

Clearly the balls do not cover . Given , the vector will be a distance away from each vector of the form where is a point on the integer lattice. So for it won't belong to any ball of the form .

Would the balls work? Why? How would the condition be changed?
 
  • #12
forgive me, I have mostly lost interest in grading gamelins homework. but I would say you should change e|ji| to e|ji|/(sqrt(n), and assume e<1 and b an integer.
 
  • #13
psie said:
I appreciate your solution, but how would one amend the intended proof?

Clearly the balls do not cover . Given , the vector will be a distance away from each vector of the form where is a point on the integer lattice. So for it won't belong to any ball of the form .

Would the balls work? Why? How would the condition be changed?
In this case, I think you should have focused on getting the required inequality first. That is:
PeroK said:
I was looking at a simple approach, where for every we can find a point in the lattice, such that:
Where is the linear lattice spacing. Then:
In fact, it seems that for any lattice, we can always find where we have equality. That suggests to me that we need the linear spacing to depend on .
Let and choose as the lattice spacing. Now, choose integer such that . We have our set of numbers:
And define the () lattice points as:
And we see that:
And, therefore, the finite set of open ballls covers , as required.

Note that if and when a proof is wrong, you may be able to patch it up, but sometimes you have to redo the whole thing and introduce some new ideas.
 
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  • #14
And notice that, since e<1 and b is an integer, that e/sqrt(n) < b, so taking all ji such that e|ji|/(sqrt(n) ≤ 2b, we do get for |ji| maximal = N, that indeed Ne/sqrt(n) = Ns ≥ 2b-e/sqrt(n) > b, as PeroK shows is what we want. I.e. so that we have enough lattice points to cover the cube.
 
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  • #15
I appreciate both of your solutions, and I'm sorry to drag this out. Upon further thought, I think the authors of the text are correct in claiming that is totally bounded by balls of the form , where with (note the strict inequality), provided we equip with the maximum metric. Here's my attempt at deriving the condition rather than .

If , then we can simply take and this will contain and clearly satisfy the condition.

For , observe that any point will be a distance less than from an integral lattice point. Consequently, any point in is less than a distance from for some . In particular, each belongs to some ball as ranges in . We want to pick only those such that .

Let be such that is in and in for some , then Now, contemplating the last statement, the interval will not contain points of if or . So negating this, the interval will contain points of if and , i.e. , and since we had , we get the desired condition, but with strict inequality, that is
 
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  • #16
If we wanted an even sharper bound, we could notice that the distance from a point to an integral lattice point is actually less than or equal to . We would then get
 
  • #17
note: it was mentioned in post #9 that using the max norm would finesse changing the radius of the balls, by changing balls into cubes.
 
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