Existence of sequences whose limit is infimum

Question

Im trying to understand the proofs of this question or this.

Let $(E,d)$ be a metric space. Let $A$, $B$ be two disjoint non-empty subsets of $E$ with $A$ compact and $B$ closed. Show that $dist(A,B):=\inf\{d(x,y): x\in A, y\in B\}>0$

Poof begins supposing $dist(A,B)=0$. Then it implies that there are of sequences $\{a_n\}$ in A and $\{b_n\}$ in B, with $d(a_n,b_n) \rightarrow 0$.

I really don't understand why is that. Could someone explain to me?

score 1 · Accepted Answer · answered Jul 08 '19 at 21:40

1

hint

Let $E=\{d(a,b) \; : \; (a,b)\in A\times B\}$.

then $d(A,B)=\inf E$.

the characterisation of the infimum gives

$$\forall \epsilon>0 \;\; \exists e\in E \; : \; d(A,B)\le e<d(A,B)+\epsilon$$

$$\iff$$ $$\forall n\in \Bbb N \; \exists e_n\in E \; : \; 0\le e_n< \frac{1}{n+1}$$

thus $$\lim_{n\to+\infty}e_n=0$$ but the existence of $e_n$ is equivalent to the existence of $(a_n,b_n)\in A\times B$ such that $e_n=d(a_n,b_n)$.

answered Jul 08 '19 at 21:40

hamam_Abdallah

62,951

As far as I know, the existence of such $e \in E$ is valid for real numbers. You used it because the image of a metric is real? – Celine Harumi Jul 08 '19 at 21:45
@Danmat Yes E is a subset of $\Bbb R^+$. – hamam_Abdallah Jul 08 '19 at 21:46
Wow I completely understand it now. You constructed a sequence in the image of $d$ in the form $d(a_n,b_n)$ for each $n$, and show that this converges in $\mathbb{R_+}$ – Celine Harumi Jul 08 '19 at 21:48
@Danmat Thanks a lot . You made me happy. Thanks once again. – hamam_Abdallah Jul 08 '19 at 22:57

Existence of sequences whose limit is infimum

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