Verifying an inequality if $A \subset \Bbb{C}$ is compact?

Question

Suppose $A,B$ are subsets of the complex plane $\Bbb{C}$ with $A$ compact

Then there exists a point $a \in A$ such that $\forall x \in A, y \in B$ there exists $b \in B$ such that $|a-b| \leq |x-y|$.

I am thinking whether the above statement is true? $A$ is compact in $\Bbb{C}$ means that $A$ is closed and bounded, trying to think $\Bbb{C}$ similar to an $\Bbb{R}^2$ plane, but I cannot think what the inequality means? any motivation or visualization of the inequality and is that even true?

Answer 1

The map $A\rightarrow \mathbb{R}^+, x\mapsto d(x,B)$ is continuous on the compact set $A$.

Here $d(x,B)=\inf\{|x-y|, \, y\in B\}$. Then there exists $a\in A$ which minimizes this map $i.e.$ $d(a,B)=d(A,B)$.

Then for every $x\in A$ and $y\in B$ we have $|x-y|\geq d(A,B) = d(a,B)=|a-b|$ for some $b\in \bar{B}$.

Answer 2

Let us define $d(x,B) = \inf_{z\in B} d(x,z)$. From the triangle inequality $d(x,z)\leq d(y,z)+d(x,y)$, by taking infimum over $z\in B$, we know that $x\mapsto d(x,B)$ is Lipschitz continuous.

Now, assume to the contrary that for every $a\in A$, there exists $x\in A$ and $y\in B$ such that for all $b\in B$ $$ d(a,b)>d(x,y) \tag{*} $$ holds. Choose $a \in \text{argmin}_{p\in A} d(p,B)$. Then $(*)$ implies that $$ \min_{p\in A} d(p,B) = d(x,y). $$ By $(*)$, there exists $x'\in A$, $y'\in B$ such that $$ \min_{p\in A} d(p,B) = d(x,y)>d(x',y')\ge d(x',B). $$ This contradict $x'\in A$. Hence we argue by contradiction that the statement is true.

Answer 3

The statement is true.

Let $M = \inf\{|a-b| : a \in A, b \in B\}$. First of all, suppose that the infimum is achieved, i.e. there exist $a \in A$ and $b \in B$ such that $|a-b| = M$. Then it is clear that for every $x \in A, y \in B$, we have $|a-b| \leq |x-y|$, so the condition is satisfied.

Conversely, suppose the infimum isn't satisfied. Then there exist$^1$ sequences $(a_n)_n \subseteq A$ and $(b_n)_n \subseteq B$ such that $|a_n - b_n| \rightarrow M$. Now, since $A$ is compact, it is sequentially compact, so we can extract a subsequence $(a_{n_k})_k$ of $(a_n)_n$ which converges to some $a \in A$. We claim this $a$ satisfies the condition.

Let $x \in A$, $y \in B$. We know that the infimum $M$ is not achieved, so we have $|x-y|>M$. Let $\epsilon = |x-y|-M>0$. By definition of convergence$^2$, we can pick $K,N \in \mathbb{N}$ such that:

\begin{align*} &\forall k \geq K, |a_{n_k} - a| < \epsilon/2 \\ &\forall n \geq N, |a_n - b_n| < \epsilon/2 + M \end{align*}

Then we can pick some $k$ such that $k \geq K$ and $n_k \geq N$, and we get:

\begin{align*} |a - b_{n_k}| \leq |a - a_{n_k}| + |a_{n_k} - b_{n_k}| < \epsilon/2 + \epsilon/2 + M = \epsilon + M = |x-y| \end{align*}

And thus choosing $b_{n_k}$ satisfies the desired condition. $\blacksquare$

$^1$Assuming $A$ and $B$ are non-empty. To see this, note that by definition of the infimum, for every $n \in \mathbb{N}$, there must exist $a_n \in A$ and $b_n \in B$ such that $|a_n - b_n| < M + 1/n$. And, naturally, $|a_n - b_n| \geq M$.

$^2$In the case of $|a_n - b_n| \rightarrow M$, we can remove one set of absolute values since we know $|a_n - b_n| > M$ for every $n$.