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Prove that the following are equivalent: a) A is bounded, b) A is in a closed ball

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Prove that the following are equivalent: a) A is bounded, b) A is "in" a closed ball

Homework Statement


The full problem is:
Let M be a metric space an A[itex]\subseteq[/itex]M be any subset. Prove that the following are equivalent:
a)A is bounded.
b)A is contained in some closed ball
c)A is contained in some open ball.

I only want help going from A to B, but maybe a little guidance from B to C or A to B--and I will attempt to prove the opposite way.


Homework Equations


Book definitions:
A is bounded if [itex]\exists[/itex]R≥0 s.t. d(x,y)≤R [itex]\forall[/itex] x,y[itex]\in[/itex]A
If a is a nonempty bounded subset of M, the diameter of A is diam(A) = sup{d(x,y):x,y[itex]\in[/itex]A}
For any x[itex]\in[/itex]M and r>0, the closed ball of radius r around x is [itex]\overline{B}[/itex]r(x)={y[itex]\in[/itex]M:d(y,x)≤R}


The Attempt at a Solution


My first thoughts are:
(=>) A to B
Spse A is bounded.
Let R = diam(A)
[itex]\forall[/itex]x1,x2[itex]\in[/itex]A, d(x1,x2)≤R
Thus, [itex]\forall[/itex]x[itex]\in[/itex]A, [itex]\exists[/itex]y[itex]\in[/itex]M s.t. d(y,x)≤R
Let [itex]\overline{B}[/itex]r(x)={y[itex]\in[/itex]M:d(y,x)≤R} be the arbitrary union of y's.
Thus, [itex]\forall[/itex]x[itex]\in[/itex]A, x[itex]\in[/itex][itex]\overline{B}[/itex]r(x)
Thus, A[itex]\subseteq[/itex][itex]\overline{B}[/itex]r(x)
 

Answers and Replies

  • #2
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You're studying Lee's book, aren't you? :smile:

My first thoughts are:
(=>) A to B
Spse A is bounded.
Let R = diam(A)
[itex]\forall[/itex]x1,x2[itex]\in[/itex]A, d(x1,x2)≤R
Thus, [itex]\forall[/itex]x[itex]\in[/itex]A, [itex]\exists[/itex]y[itex]\in[/itex]M s.t. d(y,x)≤R
Let [itex]\overline{B}[/itex]r(x)={y[itex]\in[/itex]M:d(y,x)≤R} be the arbitrary union of y's.
Thus, [itex]\forall[/itex]x[itex]\in[/itex]A, x[itex]\in[/itex][itex]\overline{B}[/itex]r(x)
Thus, A[itex]\subseteq[/itex][itex]\overline{B}[/itex]r(x)
No, this can't be correct. You have shown that [itex]x\in B_r(x)[/itex]. But here you find for each x, a ball that contains x. So you have a ball for each x in A. You don't want that. You want only one ball that contains all the elements in A. So you want to find an [itex]B_r(x)[/itex] such that [itex]y\in B_r(x)[/itex] for all y in A.

Now, what if you just take r like you did before, and take x arbitrary (but fixed)??
 
  • #3
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Lol, I sure am! Brand new edition! I see what you are saying!
This time I will fix x (since by def of diameter, all x's will be inside that).

Spse A is bounded,
Let R = diam(A)
Let x[itex]\in[/itex]A, [itex]\exists[/itex]y[itex]\in[/itex]M s.t. d(x,y)≤R
Let [itex]\bar{B}[/itex]r(x) = {y[itex]\in[/itex]M:d(y,x)≤R}
Thus [itex]\forall[/itex]x[itex]\in[/itex]A, x[itex]\in[/itex][itex]\bar{B}[/itex]r(x) (since [itex]\forall[/itex]x1,x2[itex]\in[/itex]A, d(x1,x2)≤R)
Thus, A[itex]\subseteq[/itex][itex]\bar{B}[/itex]r(x)
 
  • #4
22,097
3,280


Lol, I sure am! Brand new edition! I see what you are saying!
This time I will fix x (since by def of diameter, all x's will be inside that).

Spse A is bounded,
Let R = diam(A)
Let x[itex]\in[/itex]A, [itex]\exists[/itex]y[itex]\in[/itex]M s.t. d(x,y)≤R
Let [itex]\bar{B}[/itex]r(x) = {y[itex]\in[/itex]M:d(y,x)≤R}
Thus [itex]\forall[/itex]x[itex]\in[/itex]A, x[itex]\in[/itex][itex]\bar{B}[/itex]r(x) (since [itex]\forall[/itex]x1,x2[itex]\in[/itex]A, d(x1,x2)≤R)
Thus, A[itex]\subseteq[/itex][itex]\bar{B}[/itex]r(x)
No, this is just the exact same proof.
What you want is to fix [itex]x_0[/itex] and R, and prove that
[itex]\forall y\in A:~y\in \overline{B}_R(x_0)[/itex].
 
  • #5
145
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Hmm, I am having a bit of trouble seeing it.
To be sure, are you saying that the y's need to be in A or can they just be within M and outside of A? Does that make sense?
 
  • #6
22,097
3,280


You want to prove that all elements in A are in a closed ball. So I pick a closed ball [itex]\overline{B}_R(x_0)[/itex] (with our previous R and [itex]x_0[/itex]).
Now, I need to prove that A is a subset of this closed ball.

So I need to prove that for all y in A holds that [itex]d(y,x_0)\leq R[/itex].
 
  • #7
145
3


I believe that I understand it a bit better. I may have just been fudging up my material a bit.
For my personal clarity, I want y to be the point within M that satisfy the inequality. We will then label points in x as x1, etc.

Spse A is bounded
Let R = diam(A)
Let x0[itex]\in[/itex]A
[itex]\exists[/itex]y[itex]\in[/itex]M s.t. d(x0,y)≤R
Let [itex]\bar{B}[/itex]R(x0) = {y[itex]\in[/itex]M : d(x0,y)≤R}
Let x1[itex]\in[/itex]A,
since d(x0,x1)≤R, x1[itex]\in[/itex][itex]\bar{B}[/itex]R(x0)
Thus, A[itex]\subseteq[/itex][itex]\bar{B}[/itex]R(x0)

To me this makes sense, by fixing the ball of length R around x0, and because R is = diam (A), we know that if an element is in A then the distance between it and x0 must also be less or equal. Thus, it is within the "area" of the ball.

How's that?

Thank you so much for your help!
 
  • #8
22,097
3,280


That's better.

For my personal clarity, I want y to be the point within M that satisfy the inequality.
The point is that all points in M satisfy the inequality.

Spse A is bounded
Let R = diam(A)
Let x0[itex]\in[/itex]A
This is ok.

[itex]\exists[/itex]y[itex]\in[/itex]M s.t. d(x0,y)≤R
This line is unnecessary.

Let [itex]\bar{B}[/itex]R(x0) = {y[itex]\in[/itex]M : d(x0,y)≤R}
Let x1[itex]\in[/itex]A,
since d(x0,x1)≤R, x1[itex]\in[/itex][itex]\bar{B}[/itex]R(x0)
Thus, A[itex]\subseteq[/itex][itex]\bar{B}[/itex]R(x0)
This is ok.
 
  • #9
145
3


Thank you very much! I see what I was doing wrong now. I love this site! lol
 

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