show that A is bounded iff diam(A) is finite

We define the diameter of a nonempty subset A of M by { }. Show that A is bounded if and only if diam(A) is finite.

this is seems obvious, if diam(A) is not finite, then the distance from a to b can be infinite, which says that A is not bounded.

However I cannot think of an official proof of this, can anyone help?

Re: show that A is bounded iff diam(A) is finite

Quote:

Originally Posted by

**wopashui** We define the diameter of a nonempty subset A of M by

{

}. Show that A is bounded if and only if diam(A) is finite.

I agree with you here. It seems that you are being asked to prove a definition.

So I guess there is some other definition of a set being bounded.

What is the definition of a bounded set in your text?

Re: show that A is bounded iff diam(A) is finite

A is bounded if there is some and some constant C<infinity s.t for all

So if We know A is bounded, we have and which implies

so this shows that diam(A) is less than infinity, hence its finite. Is this correct?

How about prove from the other direction?

Re: show that A is bounded iff diam(A) is finite

Quote:

Originally Posted by

**wopashui** A is bounded if there is some

and some constant C<infinity s.t

for all

So if We know A is bounded, we have

and

which implies

so this shows that diam(A) is less than infinity, hence its finite. Is this correct?

How about prove from the other direction?

Well OK. Suppose that

Let then .

What can you with that?

Re: show that A is bounded iff diam(A) is finite

Quote:

Originally Posted by

**Plato** Well OK. Suppose that

Let

then

.

What can you with that?

what is here? Is it the diamater of A? Should we let ?

Re: show that A is bounded iff diam(A) is finite

Re: show that A is bounded iff diam(A) is finite

Re: show that A is bounded iff diam(A) is finite

Quote:

Originally Posted by

**wopashui** this is contridiction since

=

should be less than

right?

That is correct.

Here we would have

Re: show that A is bounded iff diam(A) is finite

Re: show that A is bounded iff diam(A) is finite

Quote:

Originally Posted by

**wopashui** but how did you get

less than d(a,b), if we wanna prove by contridcition

Don't you see that so some pair that have the property ?

Re: show that A is bounded iff diam(A) is finite

yes, I see that, but i just not sure how you got the last inequality, since by definition x1 x2 cannot both be an element of A, one of them must be in M, thats wy i have ,

Re: show that A is bounded iff diam(A) is finite

Re: show that A is bounded iff diam(A) is finite

ok , isee now, I forgot to use the face we have proved that if A is bounded, then d(a,b)<= C for some C

thank you