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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. –  mtiano Feb 1 at 16:06
    
You can also prove it directly. –  Asaf Karagila Feb 1 at 16:06
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There's really no need for contradiction: You're taking the supremum over a smaller set, so you don't get a larger result. –  T. Bongers Feb 1 at 16:07
    
I am having difficulty proving the answer is yes. Any help? thanks –  user124140 Feb 1 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 at 17:29
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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 at 16:20
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