Dtyjlui

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?

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

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

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

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

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

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 $n$.