Geometric Interpretation of "$A,B\subseteq \mathbb{C}$, there exists a point $a∈A$ such that $∀x∈A,y∈B $ there exists $b∈B$ such that $|a−b|≤|x−y|.$

Question

Suppose $A,B$ are subsets of the complex plane $C$ with $A$ compact.

Then there exists a point $a∈A$ such that $∀x∈A,\ y∈B $ there exists $b∈B$ such that $|a−b|≤|x−y|$.

This is not a duplicate of Show that there exists a point $a \in A$ such that for all $x \in A$ and $y \in B$, there exists $b \in B$ such that $|a - b| \le |x - y|$.

I have proved it. I just can't see it geometrically. Can some one help in visualizing what is happening here.

Thanks.

Answer 1

It's just saying that there is a shortest distance between the two sets, specified by the distance between the points $a$ and $b$. In other words, if you were to choose any two points $x\in A$ and $y \in B$, that these two points will be further apart than the distance between $a$ and $b$.

(Of course, if the two sets $A$ and $B$ intersect, then you can just choose $a = b \in A\cap B$, then $|a - b| = 0$. )