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

**wopashui** · October 16th 2011, 02:01 PM

We define the diameter of a nonempty subset A of M by $diam(A)=sup$ { $d(a,b): a,b \in A$ }. 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?

**Plato** · October 16th 2011, 02:29 PM

Originally Posted by wopashui

We define the diameter of a nonempty subset A of M by $diam(A)=sup$ { $d(a,b): a,b \in A$ }. 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?

**wopashui** · October 16th 2011, 02:36 PM

A is bounded if there is some $x\inM$ and some constant C<infinity s.t $d(a,x)<=C$ for all $a\inA$

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

How about prove from the other direction?

**Plato** · October 16th 2011, 04:16 PM

Originally Posted by wopashui

A is bounded if there is some $x\inM$ and some constant C<infinity s.t $d(a,x)<=C$ for all $a\inA$ So if We know A is bounded, we have $d(a,x)<=C$ and $d(b,x)<=C$ which implies $d(a,b)<= d(a,x)+d(x,b)<=C+C=2C< infinity$
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 $\delta(A)<\infty$
Let $x_0\in A$ then $\left( {\forall x \in A} \right)\left[ {d(x_0 ,x) \le \delta (A)} \right]$ .
What can you with that?

**wopashui** · October 16th 2011, 06:02 PM

Originally Posted by Plato

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

what is $\delta(A)$ here? Is it the diamater of A? Should we let $x_0 \in$ $M$ ?

**Plato** · October 17th 2011, 03:07 AM

Originally Posted by wopashui

what is $\delta(A)$ here? Is it the diamater of A? Should we let $x_0 \in$ $M$ ?

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

**wopashui** · October 17th 2011, 02:11 PM

Originally Posted by Plato

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

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

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

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

thus $d(a,b) <= d(a,x)+d(b,x) >2C$ , but we can't say that $d(a,b) > 2C$ from this inequality

**Plato** · October 17th 2011, 02:19 PM

Originally Posted by wopashui

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

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

**wopashui** · October 17th 2011, 02:31 PM

Originally Posted by Plato

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

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

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

thus $d(a,b) <= d(a,x)+d(b,x) >2C$ , but we can't say that $d(a,b) > 2C$ from this inequality

**Plato** · October 17th 2011, 02:44 PM

Originally Posted by wopashui

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

Don't you see that $C=\delta (A)+c>0$ so some pair $x_1~\&~x_2$ that have the property $\delta (A)+c\le d(x_1,x_x_2)$ ?

**wopashui** · October 17th 2011, 02:58 PM

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 $d(a,b) <= d(a,x)+d(b,x) >2C$ ,

**Plato** · October 17th 2011, 03:10 PM

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 $d(a,b) <= d(a,x)+d(b,x) >2C$ ,

You have a problem with negations.

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

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

So you see that $x_1\in A~\&~x_2\in A$ .

**wopashui** · October 17th 2011, 03:59 PM

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

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