Is A Bounded by a Closed Ball?

In summary: I know I'll be back here soon with more questions.In summary, To prove that A is bounded, we choose an arbitrary element x0∈A and create a closed ball X¯r(x0) with radius R=diam(A). Then, for any other element x1∈A, we know that d(x0,x1)≤R, and therefore x1∈X¯r(x0). Thus, A is contained in the closed ball X¯r(x0), proving that A is bounded.
  • #1
Hodgey8806
145
3
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)
 
Physics news on Phys.org
  • #2


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

Hodgey8806 said:
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


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


Hodgey8806 said:
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


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


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


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


That's better.

Hodgey8806 said:
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


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

What does it mean for a set A to be bounded?

Boundedness of a set A means that there exists a finite number M such that the distance between any two elements in A is less than or equal to M. In other words, the elements in A are contained within a specific range or interval.

What is a closed ball?

A closed ball is a set of points that are all within a certain distance from a given point, known as the center of the ball. The boundary of a closed ball is included in the set, meaning that the distance between the boundary points and the center point is equal to the radius of the ball.

How are bounded sets and closed balls related?

A set A is considered bounded if and only if it is contained within a closed ball with a finite radius. This means that all the elements in A are within a certain distance from a given point, just like the definition of a closed ball.

Can a set be bounded without being in a closed ball?

No, a set cannot be bounded without being in a closed ball. If a set is bounded, it means that the elements are contained within a specific range or interval, and this can only happen if the set is within a closed ball with a finite radius.

How can you prove that a set A is in a closed ball if it is bounded?

To prove that a set A is in a closed ball, you must show that the distance between any two elements in A is less than or equal to the radius of the closed ball. This can be done by assuming that A is bounded and using the definition of boundedness to show that it is contained within a closed ball with a finite radius.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
249
  • Calculus and Beyond Homework Help
Replies
1
Views
454
  • Calculus and Beyond Homework Help
Replies
3
Views
508
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
804
  • Calculus and Beyond Homework Help
Replies
2
Views
252
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
548
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
736
Back
Top