Two disjoint closed sets $A,B$ with $B$ compact, show $d(A,B) > 0$ Verify my proof

Question

Two disjoint closed sets $E,F$ with $E$ compact, show $d(E,F) > 0$.

So, with compactness we get a few things, that every sequence has a convergent subsequence and we can use the extreme value theorem. I saw a proof on this site where they use convergence of subsequences, but I'd like to see one using the extreme value theorem, as I'm sure is possible. Here is what I've came up with so far.

Proof: Define a function $f: E \rightarrow \mathbb{R}$ by $f(e)=d(e,F)$. One can check that this function is continuous.

Since this is a continuous function on a compact set $E$, we can apply the extreme value theorem. Thus $x \in E$ s.t. $f(x) \leq f(e)$ $\forall e \in E$.

Since $E \cap F = \emptyset$ and $x \in E$ we have $d(x,F)>0$

Now i'm a bit stuck... Can anyone help out?... I have yet to use the fact that $F$ is closed.

Answer 1

$F$ is closed, so $x \notin F$ implies $d(x,F)>0$: As $x \in X\setminus F$ and the latter set is closed, there is some $r>0$ such that $B(x,r) \subseteq X\setminus F$, which implies that when $y \in F$, $d(x,y) \ge r$.

So $d(x,F)=\inf\{d(x,y): y \in F\} \ge r >0$.

So after you applied that $f(x)=d(x,F)$ has a minimum $p \in E$ by compactness of $E$, the closedness of $F$ and the fact that $p \in E$ implies $p \notin F$ by disjointness, we have that $d(A,B) \ge f(p) >0$, finishing the proof.

Answer 2

Prove by contradiction. Suppose $d(E,F)=0$. Then there exist $(e_n) \subset E,(f_n) \subset F$ with $d(e_n,f_n) \to 0$. $(e_n)$ has a subsequence $(e_{n_k})$ converging to some point $e \in E$. Now use triangle inequality to show that $(f_{n_k})$ also tends to $e$. But $F$ is closed so the limit of $(f_{n_k})$ must belong to $F$. Thus we get the contradiction that $e \in E \cap F$.

Alternatively, use your argument and just observe that $d(E,F)=\inf \{d(e,F): e \in E\}=f(x)=d(x,F) >0$.