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I tried to prove if $A \subseteq B$ then $\sup_{a,a' \in A} d(a,a') \le \sup_{b,b' \in B} d(b,b')$ where $d$ is some metric and $A,B$ are some sets in some metric space. But both attempts fail even though the statement seems obvious to me.

One attempt: Let $A \subseteq B$. Then there is a sequence $(a_n, a'_n) $ in $A \times A$ with $d(a_n, a'_n) \to \sup_{a,a' \in A} d(a,a')$. (The details of why this is escapes me. It seems to be that one must assume $X \times X$ to be complete with respect to $d$?) Since $A \subseteq B$ this sequence is also a sequence in $B$ converging to $\sup_{a,a' \in A} d(a,a')$. From this somehow one should be able to conclude that $\sup d(a,a') \le \sup d(b,b')$. But how?

Another attempt: Assume there were $a,a' \in A$ with $d(a,a') > d(b,b')$ for all $b,b' \in B$. Then $a,a'$ are also in $B$ which contradicts $>$. Therefore it must holds that for every $a,a' \in A$ there are $b,b' \in B$ with $d(a,a') \le d(b,b')$. Here I want to say: Then it follows that there are $b, b' \in B$ with $\sup_{a,a' \in A} d(a,a') \le d(b,b')$. But I'm not sure it holds.

How to prove it or how to fix my attempts?

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The definition of $\sup S$ is that it is an upper bound for $S \subset \mathbb{R}$, and for any $\epsilon>0$, $[\sup S -\epsilon, \infty) \cap S \neq \emptyset$. You don't need completeness. Once you have the sequence $(a_n,a_n')$ above (which exists by definition), then the result is immediate, since $(a_n,a_n') \in A \subset B$. –  copper.hat Feb 13 '13 at 19:22

2 Answers 2

up vote 1 down vote accepted

If $S \subset T$ and $f$ is a function defined on $T$, then $\sup_{x \in S} f(x) \leq \sup_{x \in T} f(x)$.

To see this, note that $f(x' ) \leq \sup_{x \in T} f(x)$ for all $x' \in T$. Hence $f(x' ) \leq \sup_{x \in T} f(x)$ for all $x' \in S$, and so $\sup_{x' \in S} f(x') \leq \sup_{x \in T} f(x)$.

Now let $T = B \times B$, and $S = A \times A$; by assumption $S \subset T$. Let $x=(x_1, x_2) \in T$, and define $f(x) = d(x_1,x_2)$. The desired result follows from this.

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Thank you, that is very neat! Can any of my two approaches be fixed? –  mathvisitor Feb 13 '13 at 19:17

It is much easier. Let $$ \mathcal{A}=\{d(x,y):x,y\in A\}\text{ and }\mathcal{B}=\{d(x,y):x,y\in B\}. $$ Since $A\subset B$, it is clear that $\mathcal{A}\subset \mathcal{B}$, and then $$ \sup\mathcal{A}\le\sup\mathcal{B}. $$

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Thank you! Can any of my two approaches be fixed? –  mathvisitor Feb 13 '13 at 19:17

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