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Suppose $(X,d)$ is a metric space, for a nonempty subset $A$ of $X$, define

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

$A$ is bounded if $\delta(A) < \infty $. If $A$, $B$ are bounded, does it follow that $A \cup B$ is bounded ? I know that $A \cap B$ is bounded since $A \cap B \subseteq B$, and the fact that $A \subseteq B \implies \delta(A) \leq \delta(B) $. But how about the $\cup$ ?

thanks

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Yes. It's bounded by $\delta(A)+\delta(B)+c$, where $c$ is the distance between a fixed point of $A$ and a fixed point of $B$. –  David Mitra Feb 2 at 10:20

2 Answers 2

up vote 0 down vote accepted

And, in order to prove David Mitra's answer, you should:

$$ \mathrm{sup}_{x\in A,y\in B} \left\{ x + y\right\} \leq \mathrm{sup}_{x\in A} \left\{ x \right\} + \mathrm{sup}_{y\in B} \left\{ y\right\} \ . $$

EDIT. And the second inequality comes from the fact that, for all $x\in A$ and all $y\in B$, you have

$$ x + y \leq \mathrm{sup}_{x\in A} \left\{ x \right\} + \mathrm{sup}_{y\in B} \left\{ y\right\} \ . $$

Now, apply $\mathrm{sup}_{x\in A,y\in B}$ to both sides of this inequality and you are done.

Exercise. Why don't you have, in general, an equality

$$ \mathrm{sup}_{x\in A,y\in B} \left\{ x + y\right\} = \mathrm{sup}_{x\in A} \left\{ x \right\} + \mathrm{sup}_{y\in B} \left\{ y\right\} \quad \text{?} $$

:-)

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can you show your second bullet? It is hard for me to convince about it so easily. thanks –  user124140 Feb 2 at 10:31
    
I've made an edit to my answer. Hope it helps. –  a.r. Feb 2 at 10:42

Let's $a\in A$ and $b\in B$ be fixed. Take any $x\in A , y\in B$, then $$d(x,y) \leq d(x,a ) +d(a,b) +d(b,y) \leq \delta (A) +d(a,b) +\delta (B)$$ thus $A\cup B$ is bounded.

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