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I come to know about this problem while reading What is the best way to define the diameter of the empty subset of a metric space? .I've already proved this but i wanted to prove $$ \operatorname{diam}(A\cup B)\le \operatorname{diam}A+\operatorname{diam}B+\operatorname{dist}(A,B) $$ in another way .But i got stuck in between.

Since ,$A \subseteq A\cup B$ & $B \subseteq A\cup B \implies $ $\operatorname{diam}{A} \le \operatorname{diam}{(A\cup B)}$ & $\operatorname{diam}{(B)} \le \operatorname{diam}{(A\cup B)} \implies\operatorname{diam}{(A\cup B)} \le \frac{\operatorname{diam}A+\operatorname{diam}B}{2} $.

Assuming that i'm on the right track,i tried to proceed further but not getting any idea.

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Styles
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  • Can you explain how you got that $$\text{diam} \left(A\cup B\right) \leq \frac{\text{diam} A + \text{diam} B}{2}$$ – Ariel Bereslavsky Sep 24 '16 at 13:38
  • @ArielBereslavsky:I've just added the two inequations. – Styles Sep 24 '16 at 13:43
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    Your last inequality is false, take for instance $A={0,1}$ and $B={2,3}$. Shouldn't it be the other way around? – Olivier Moschetta Sep 24 '16 at 13:43
  • You somehow add them backwards. Check your math again. Suppose that $\text{diam} (A) = \text{diam} (B)$ then – Ariel Bereslavsky Sep 24 '16 at 13:49
  • Possible duplicate: http://math.stackexchange.com/questions/932579/show-that-operatornamediama-cup-b-le-operatornamediama-operatorname/932592#932592 – Quang Hoang Sep 24 '16 at 14:00
  • @OlivierMoschetta:Thanks for pointing this.Actually,I've read a theorem which states that-If $S$ & $T$ are two subsets of a metric space $(X,d)$,then if $S \subset T$ then $\operatorname{diam}{(S)} \le \operatorname{diam}{(T)}$. – Styles Sep 24 '16 at 15:37
  • @OlivierMoschetta:From this we can conclude that that the inequality in OP holds if if $A$ & $B$ are subsets of metric space NOT of their unions.Thanks for this nice counterexample. – Styles Sep 24 '16 at 15:40

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As already mentioned in comments, your idea does not work.

For $a,a'\in A$, $b,b'\in B$ we have: $$ d(a,b)\leq d(a,a')+d(a',b')+d(b',b)\leq {\rm diam}(A)+ d(a',b') + {\rm diam} (B) $$ Now take inf on the RHS over $a'\in A$, $b'\in B$.

H. H. Rugh
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