I'm looking at the following math problem:
If $C$ is compact and $D$ is closed, then there are points $c \in C$ and $d \in D$ such that $d(C, D) = |c - d|.$
Here is an (unofficial) solution to the problem:
We have that $d:\mathbb R\to \mathbb R:\ x\mapsto d(x,D)$ is comtinuous. Now, $C$ is compact, so there is a $c\in C$ and a real number $d'$ such that $d(c,D)=d'$ and $(c,d')$ is a minimum for $d$ on $C$.
So by definition of $d$, there is a sequence $(x_n)\subseteq D$ such that
$\tag1 d(c,x_n)=|c-x_n|<d'+1/n,$
$(1)$ implies that $(x_n)$ is a bounded sequence of numbers in $D,$ so that $(x_n)$ has a convergent subsequence $(x_{n_k})$. That is $x_{n_k}\to d''\in D$ (because $D$ is closed).
We conclude that $d(c,x_{n_k})\to d(c,d'')$ (because $d$ is continuous), and that (from $(1)$), $d(c,d'')\le d'$. But, equality must hold because $d'$ is a minimum, so in fact, $d(c,d'')=d'$.
But, I don't understand why
Now, $C$ is compact, so there is a $c\in C$ and a real number $d'$ such that $d(c,D)=d'$ and $(c,d')$ is a minimum for $d$ on $C$.
is true. Can someone please explain? Is this solution correct?
$d(C,D)$ is the minimum value of $d(x,D)$ as $x$ ranges over $C$. In other words it is the minimum of the the continuous function $x \to d(x,D)$ on the compact set $C$. Every continuous function on a compact set attains its minimum value. [ To prove that $x \to d(x,D)$ is continuous you can use triangle inequlaity to verify that $|d(x,D)-d(x',D)|\leq d(x,x')$].