Is A Bounded by a Closed Ball?

[Hodgey8806]

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\subseteqM 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 \existsR≥0 s.t. d(x,y)≤R \forall x,y\inA
If a is a nonempty bounded subset of M, the diameter of A is diam(A) = sup{d(x,y):x,y\inA}
For any x\inM and r>0, the closed ball of radius r around x is \overline{B}_r(x)={y\inM:d(y,x)≤R}

The Attempt at a Solution

My first thoughts are:
(=>) A to B
Spse A is bounded.
Let R = diam(A)
\forallx₁,x₂\inA, d(x₁,x₂)≤R
Thus, \forallx\inA, \existsy\inM s.t. d(y,x)≤R
Let \overline{B}_r(x)={y\inM:d(y,x)≤R} be the arbitrary union of y's.
Thus, \forallx\inA, x\in\overline{B}_r(x)
Thus, A\subseteq\overline{B}_r(x)

 

[micromass]

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

My first thoughts are:
(=>) A to B
Spse A is bounded.
Let R = diam(A)
\forallx₁,x₂\inA, d(x₁,x₂)≤R
Thus, \forallx\inA, \existsy\inM s.t. d(y,x)≤R
Let \overline{B}_r(x)={y\inM:d(y,x)≤R} be the arbitrary union of y's.
Thus, \forallx\inA, x\in\overline{B}_r(x)
Thus, A\subseteq\overline{B}_r(x)

No, this can't be correct. You have shown that x\in B_r(x). 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 B_r(x) such that y\in B_r(x) for all y in A.

Now, what if you just take r like you did before, and take x arbitrary (but fixed)??

 

[Hodgey8806]

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\inA, \existsy\inM s.t. d(x,y)≤R
Let \bar{B}_r(x) = {y\inM:d(y,x)≤R}
Thus \forallx\inA, x\in\bar{B}_r(x) (since \forallx₁,x₂\inA, d(x₁,x₂)≤R)
Thus, A\subseteq\bar{B}_r(x)

 

[micromass]

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\inA, \existsy\inM s.t. d(x,y)≤R
Let \bar{B}_r(x) = {y\inM:d(y,x)≤R}
Thus \forallx\inA, x\in\bar{B}_r(x) (since \forallx₁,x₂\inA, d(x₁,x₂)≤R)
Thus, A\subseteq\bar{B}_r(x)

No, this is just the exact same proof.
What you want is to fix x_0 and R, and prove that
\forall y\in A:~y\in \overline{B}_R(x_0).

 

[Hodgey8806]

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?

 

[micromass]

You want to prove that all elements in A are in a closed ball. So I pick a closed ball \overline{B}_R(x_0) (with our previous R and x_0).
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 d(y,x_0)\leq R.

 

[Hodgey8806]

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\inA
\existsy\inM s.t. d(x₀,y)≤R
Let \bar{B}_R(x₀) = {y\inM : d(x₀,y)≤R}
Let x₁\inA,
since d(x₀,x₁)≤R, x₁\in\bar{B}_R(x₀)
Thus, A\subseteq\bar{B}_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!

 

[micromass]

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\inA

This is ok.

\existsy\inM s.t. d(x₀,y)≤R

This line is unnecessary.

Let \bar{B}_R(x₀) = {y\inM : d(x₀,y)≤R}
Let x₁\inA,
since d(x₀,x₁)≤R, x₁\in\bar{B}_R(x₀)
Thus, A\subseteq\bar{B}_R(x₀)

This is ok.

 

[Hodgey8806]

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