Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ be a non-empty compact subset of $X$. Prove that there exist points $a,b \in A$ such that $$d(a,b) = \sup \left\{ d(x,y):x,y \in A \right\}\,.$$

The question was split into two pieces with us needing to show:

$$ | d(x,y) - d(x',y') | \leq d(x,x') + d(y,y')$$

My attempt was to say that if $A$ is compact $\exists a,b \in A$ such that there are Cauchy sequences $ \left\{ x_n \right\} , \left\{ y_n \right\} $ such that for $ \epsilon_1, \epsilon_2 > 0,\, d(x_n,a) < \epsilon_1\text{ and }d(y_n,b) < \epsilon_2 \,$.

Using this I can say:

$$ | d(x_n,y_n) - d(a,b) | \leq d(x_n,a) + d(y_n,b) < \epsilon_1 + \epsilon_2 = \epsilon_3$$

$\square$

Is this completely wrong?

share|improve this question
1  
Where do you get $a, b$ from? –  Dylan Moreland Dec 9 '11 at 7:24

1 Answer 1

That sounds a little fishy. I think what they meant you to do, or what is the quickest solution, is to note that evidently $d:A\times A\to\mathbb{R}$ is continuous (it's the restriction of the continuous map $d:X\times X\to\mathbb{R}$ to the subspace $A\times A$). But, $A\times A$ is compact (this is Tychonoff's theorem) and so by the extreme value theorem $d$ attains a maximum on $A\times A$.

share|improve this answer
1  
I agree. I think the hint is meant to help you show that the metric is continuous on $X \times X$. Just a note: proving that a product of two compact spaces is compact is not so bad. I remember Tychonoff being kind of tricky. –  Dylan Moreland Dec 9 '11 at 7:30
1  
@DylanMoreland I assume you mean that you don't need the full power of Tychonoff's theorem to prove the two-space case--indeed. But, I always try to name "googlable" results. –  Alex Youcis Dec 9 '11 at 7:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.