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I'm beginning to study metric spaces and I see this question

Consider $A$ and $B$ bounded and non-empty subsets of $M$, where $M$ is a metric space.
Show that $\operatorname{diam}(A\cup B)\le \operatorname{diam}(A)+\operatorname{diam}(B)+d(A,B)$.

Does anyone know how can I do?

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For any two points $a,b\in A\cup B$, there are three cases:

  1. $a,b\in A$, then $d(a,b)\le diam(A)$.
  2. $a,b\in B$, then $d(a,b)\le diam(B)$.
  3. $a\in A$ and $b\in B$, then for any $x\in A,y\in B$: $$d(a,b)\le d(a,x)+d(x,y)+d(y,b)\le diam(A)+d(x,y)+diam(B).$$ Now, take $\inf$ of the right hand side.
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  • $\begingroup$ And sup of the left hand side. $\endgroup$ – egreg Sep 15 '14 at 17:40

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