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Suppose $A$ is a nonempty set in a metric space $(X, d)$. Define

$$ \delta(A) = \sup_{x,y \in A} d(x,y) $$

Is it true that if $A \subseteq B$, then $\delta(A) \leq \delta(B) $??

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Yes, you can prove it by contradiction. – Wintermute Feb 1 '14 at 16:06
You can also prove it directly. – Asaf Karagila Feb 1 '14 at 16:06
There's really no need for contradiction: You're taking the supremum over a smaller set, so you don't get a larger result. – user61527 Feb 1 '14 at 16:07
I am having difficulty proving the answer is yes. Any help? thanks – user124140 Feb 1 '14 at 16:16
Beauty contest principle: IF you are sup-ing over a larger set, the sup gets bigger. If you are inf-ing over a larger set, the inf gets smaller. – ncmathsadist Feb 1 '14 at 17:29

1 Answer 1

up vote 3 down vote accepted

Recall the definition of the supremum: It's the least upper bound on a set of real numbers. Now if $C \subseteq D$ are sets of real numbers, every upper bound on $D$ is an upper bound on $C$; this proves that $\sup C \le \sup D$.

Finally, since $A \subseteq B$, it's immediate that

$$\{d(x, y) : x, y \in A\} \subseteq \{d(x, y) : x, y \in B\}$$

and by applying the previous paragraph, we're done.

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thank you very much for your time!! – user124140 Feb 1 '14 at 16:20

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