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

• Oct 16th 2011, 02:01 PM
wopashui
show that A is bounded iff diam(A) is finite
We define the diameter of a nonempty subset A of M by $\displaystyle diam(A)=sup${$\displaystyle d(a,b): a,b \in A$}. 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?
• Oct 16th 2011, 02:29 PM
Plato
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 $\displaystyle diam(A)=sup${$\displaystyle d(a,b): a,b \in A$}. 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?
• Oct 16th 2011, 02:36 PM
wopashui
Re: show that A is bounded iff diam(A) is finite
A is bounded if there is some $\displaystyle x\inM$and some constant C<infinity s.t $\displaystyle d(a,x)<=C$for all $\displaystyle a\inA$

So if We know A is bounded, we have $\displaystyle d(a,x)<=C$and $\displaystyle d(b,x)<=C$ which implies $\displaystyle d(a,b)<= d(a,x)+d(x,b)<=C+C=2C< infinity$
so this shows that diam(A) is less than infinity, hence its finite. Is this correct?

How about prove from the other direction?
• Oct 16th 2011, 04:16 PM
Plato
Re: show that A is bounded iff diam(A) is finite
Quote:

Originally Posted by wopashui
A is bounded if there is some $\displaystyle x\inM$and some constant C<infinity s.t $\displaystyle d(a,x)<=C$for all $\displaystyle a\inA$ So if We know A is bounded, we have $\displaystyle d(a,x)<=C$and $\displaystyle d(b,x)<=C$ which implies $\displaystyle d(a,b)<= d(a,x)+d(x,b)<=C+C=2C< infinity$
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 $\displaystyle \delta(A)<\infty$
Let $\displaystyle x_0\in A$ then $\displaystyle \left( {\forall x \in A} \right)\left[ {d(x_0 ,x) \le \delta (A)} \right]$.
What can you with that?
• Oct 16th 2011, 06:02 PM
wopashui
Re: show that A is bounded iff diam(A) is finite
Quote:

Originally Posted by Plato
Well OK. Suppose that $\displaystyle \delta(A)<\infty$
Let $\displaystyle x_0\in A$ then $\displaystyle \left( {\forall x \in A} \right)\left[ {d(x_0 ,x) \le \delta (A)} \right]$.
What can you with that?

what is $\displaystyle \delta(A)$ here? Is it the diamater of A? Should we let $\displaystyle x_0 \in$$\displaystyle M? • Oct 17th 2011, 03:07 AM Plato Re: show that A is bounded iff diam(A) is finite Quote: Originally Posted by wopashui what is \displaystyle \delta(A) here? Is it the diamater of A? Should we let \displaystyle x_0 \in$$\displaystyle M$?

Yes $\displaystyle \delta(A)$ stands for the diameter of $\displaystyle A.$
Suppose $\displaystyle A$ is not bounded but $\displaystyle \delta(A)<\infty$.
If $\displaystyle c>0$ then $\displaystyle \left( {\exists x_1 \in A} \right)\left( {\exists x_2 \in A} \right)\left[ {d(x_1 ,x_2 ) > \delta (A) + c} \right]$. Why is that?
• Oct 17th 2011, 02:11 PM
wopashui
Re: show that A is bounded iff diam(A) is finite
Quote:

Originally Posted by Plato
Yes $\displaystyle \delta(A)$ stands for the diameter of $\displaystyle A.$
Suppose $\displaystyle A$ is not bounded but $\displaystyle \delta(A)<\infty$.
If $\displaystyle c>0$ then $\displaystyle \left( {\exists x_1 \in A} \right)\left( {\exists x_2 \in A} \right)\left[ {d(x_1 ,x_2 ) > \delta (A) + c} \right]$. Why is that?

this is contridiction since $\displaystyle \delta (A)$ = $\displaystyle max d(x_1,x_2),$so $\displaystyle d(x_1,x_2)$should be less than $\displaystyle \delta (A)$ right?

but how did you get $\displaystyle \delta (A)$ less than d(a,b), if we wanna prove by contridcition

we have if A is not bounded, there exists $\displaystyle a,b \inA, d(a,x) > C$ for all $\displaystyle x \in M$ and $\displaystyle d(b,x) > C$for all $\displaystyle x \in M$

thus $\displaystyle d(a,b) <= d(a,x)+d(b,x) >2C$, but we can't say that $\displaystyle d(a,b) > 2C$from this inequality
• Oct 17th 2011, 02:19 PM
Plato
Re: show that A is bounded iff diam(A) is finite
Quote:

Originally Posted by wopashui
this is contridiction since $\displaystyle \delta (A)$ = $\displaystyle max d(x_1,x_2), so d(x_1,x_2)$should be less than $\displaystyle \delta (A)$ right?

That is correct.
Here we would have $\displaystyle \delta(A)+c< d(x_1,x_2)\le \delta(A)$
• Oct 17th 2011, 02:31 PM
wopashui
Re: show that A is bounded iff diam(A) is finite
Quote:

Originally Posted by Plato
That is correct.
Here we would have $\displaystyle \delta(A)+c< d(x_1,x_2)\le \delta(A)$

but how did you get $\displaystyle \delta (A)$ less than d(a,b), if we wanna prove by contridcition

we have if A is not bounded, there exists a,b $\displaystyle \in$A, d(a,x) > C for all x $\displaystyle \in$M and $\displaystyle d(b,x) > C$for all $\displaystyle x \in M$

thus $\displaystyle d(a,b) <= d(a,x)+d(b,x) >2C$, but we can't say that $\displaystyle d(a,b) > 2C$from this inequality
• Oct 17th 2011, 02:44 PM
Plato
Re: show that A is bounded iff diam(A) is finite
Quote:

Originally Posted by wopashui
but how did you get $\displaystyle \delta (A)+c$ less than d(a,b), if we wanna prove by contridcition

Don't you see that $\displaystyle C=\delta (A)+c>0$ so some pair $\displaystyle x_1~\&~x_2$ that have the property $\displaystyle \delta (A)+c\le d(x_1,x_x_2)$?
• Oct 17th 2011, 02:58 PM
wopashui
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 $\displaystyle d(a,b) <= d(a,x)+d(b,x) >2C$,
• Oct 17th 2011, 03:10 PM
Plato
Re: show that A is bounded iff diam(A) is finite
Quote:

Originally Posted by wopashui
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 $\displaystyle d(a,b) <= d(a,x)+d(b,x) >2C$,

You have a problem with negations.

Here is the definition of bounded set that you gave.
The set $\displaystyle A$ is bounded if and only if
$\displaystyle \left( {\exists C > 0} \right)\left( {\forall x \in A} \right)\left( {\forall y \in A} \right)\left[ {d(x,y) \leqslant C} \right]$.

So to say $\displaystyle A$ is not bounded if and only if
$\displaystyle \left( {\forall C > 0} \right)\left( {\exists x \in A} \right)\left( {\exists y \in A} \right)\left[ {d(x,y) > C} \right]$

So you see that $\displaystyle x_1\in A~\&~x_2\in A$.
• Oct 17th 2011, 03:59 PM
wopashui
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