I'm working out the following problem form Ahlfors' Complex Analysis text:
"Let $X$ and $Y$ be compact sets in a complete metric space $(S,d)$. Prove that there exist $x \in X,y \in Y$ such that $d(x,y)$ is a minimum."
My attempt:
Define $E:=\{d(x,y): x \in X,y \in Y \}$. We will prove that $E \subset \mathbb R$ is compact.
Firstly we will prove that $E$ is bounded:
$X$ is compact, and therefore it is bounded. That is, there exist $x_0 \in X,r_1>0$ such that $d(x,x_0)<r_1$ for all $x \in X$. $Y$ is also compact, and similarly there exist $y_0 \in Y,r_2>0$ such that $d(y,y_0)<r_2$ for all $y \in Y$.
Now, for any $d(x,y) \in E$, we have $$d(x,y) \leq d(x,x_0)+d(x_0,y_0)+d(y,y_0)<r_1+d(x_0,y_0)+r_2 =:M.$$ This proves that $E$ is bounded.
And now we will prove that $E$ is also closed:
Let $a_n=d(x_n,y_n)$ be sequence in $E$, which is convergent in $\mathbb R$. We will prove that its limit $a:=\lim_{n \to \infty} a_n$ is in $E$.
$(x_n)$ is a sequence in the compact set $X$, therefore it admits a convergent subsequence $(x_{n_k}) \to \bar{x}$. $(y_{n_k})$ is a sequence in the compact set $Y$, and therefore it admits a convergent subsequence $(y_{n_{k_l}}) \to \bar{y}$. The sequence $(x_{n_{k_l}},y_{n_{k_l}})_l$ converges to $(\bar{x},\bar{y})$ in the product space, and the continuity of the metric gives $a=\lim_{l \to \infty} a_{n_{k_l}}=\lim_{l \to \infty} d(x_{n_{k_l}},y_{n_{k_l}})=d(\bar{x},\bar{y})$. This shows that $a \in E$ as required.
In summary I have shown that $E$ is compact, and from a well-known theorem it has a minimum.
Is this proof OK? I think that the completeness of $S$ is unnecessary.
Thank you.
Best Answer
The boundedness of $E$ isn't really the problem. We only need $E$ to be bounded below, which is the case : it is the set of the values of a positive function.
The main part is to prove that $E$ is closed. And the fact that a bounded closed subset of $\mathbb{R}$ contains its minimum isn't really a famous theorem : it's only the fact that the infimum of a set is a limit point of it, and therefore contained in it if the set is closed.