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Let $(X,d)$ be a complete metric space.

$$CB(X)=\{A : A \text{ is a nonempty closed and bounded subset of }X \},$$ $$D(A,B)=\inf \{d(a,b) : a\in A , b\in B\},$$ $$\sigma (A,B)=\sup \{d(a,b) : a\in A , b\in B\},$$ $$H(A,B)=\max \{\sup_{x\in B} D(x,A) , \sup_{x\in A} D(x,B)\}.$$

Lemma: Let $A,B\in CB(X)$, and let $x\in A$. Then, for each $\alpha>0$, there exists a $y\in B$ such that \begin{equation} d(x,y)\leq H(A,B)+\alpha. \end{equation}

Question : How can we prove the lemma, and that this lemma remains valid in b-metric space?

A b-metric space means the same as metric space, with triangle inequality replaced with: $$\exists s\ge 1:\quad \forall x,y,z\in X:\quad d(x,z)\le s\big(d(x,y)+d(y,z)\big).$$

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    $\begingroup$ What's "b-metric space"? $\endgroup$
    – YCor
    Commented Sep 29, 2020 at 20:17
  • $\begingroup$ The same definition of metric space just the difference in the triangular inequality. $\endgroup$
    – Seddik Merdaci
    Commented Sep 29, 2020 at 20:28
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    $\begingroup$ Please give a complete definition $\endgroup$
    – YCor
    Commented Sep 29, 2020 at 20:28
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    $\begingroup$ Let X be a nonempty set and s≥1 be a given real number. A function d:X×X→R+ is a b-metric on X if for all x,y,z∈X the following conditions hold:\ d ( x , y ) = 0 if and only if x = y . d ( x , y ) = d ( y , x ). d ( x , z ) \leq s [ d ( x , y ) + d ( y , z ) ]. In this case, the pair (X,d) is called a b-metric space. $\endgroup$
    – Seddik Merdaci
    Commented Sep 29, 2020 at 20:45

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