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$\subseteq$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 $\exists$R≥0 s.t. d(x,y)≤R $\forall$ x,y$\in$A
If a is a nonempty bounded subset of M, the diameter of A is diam(A) = sup{d(x,y):x,y$\in$A}
For any x$\in$M and r>0, the closed ball of radius r around x is $\overline{B}$_r(x)={y$\in$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)
$\forall$x₁,x₂$\in$A, d(x₁,x₂)≤R
Thus, $\forall$x$\in$A, $\exists$y$\in$M s.t. d(y,x)≤R
Let $\overline{B}$_r(x)={y$\in$M:d(y,x)≤R} be the arbitrary union of y's.
Thus, $\forall$x$\in$A, 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)
$\forall$x₁,x₂$\in$A, d(x₁,x₂)≤R
Thus, $\forall$x$\in$A, $\exists$y$\in$M s.t. d(y,x)≤R
Let $\overline{B}$_r(x)={y$\in$M:d(y,x)≤R} be the arbitrary union of y's.
Thus, $\forall$x$\in$A, 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$\in$A, $\exists$y$\in$M s.t. d(x,y)≤R
Let $\bar{B}$_r(x) = {y$\in$M:d(y,x)≤R}
Thus $\forall$x$\in$A, x$\in$$\bar{B}$_r(x) (since $\forall$x₁,x₂$\in$A, d(x₁,x₂)≤R)
Thus, A$\subseteq$$\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$\in$A, $\exists$y$\in$M s.t. d(x,y)≤R
Let $\bar{B}$_r(x) = {y$\in$M:d(y,x)≤R}
Thus $\forall$x$\in$A, x$\in$$\bar{B}$_r(x) (since $\forall$x₁,x₂$\in$A, d(x₁,x₂)≤R)
Thus, A$\subseteq$$\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$\in$A
$\exists$y$\in$M s.t. d(x₀,y)≤R
Let $\bar{B}$_R(x₀) = {y$\in$M : d(x₀,y)≤R}
Let x₁$\in$A,
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$\in$A

This is ok.

$\exists$y$\in$M s.t. d(x₀,y)≤R

This line is unnecessary.

Let $\bar{B}$_R(x₀) = {y$\in$M : d(x₀,y)≤R}
Let x₁$\in$A,
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