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

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.

• Yes, I believe I can without much trouble, and yet even assuming this triviality my proof is not yet complete. – Mathematical Mushroom May 11 at 23:24
• Yes, I've gotten that far already, as you can see above. – Mathematical Mushroom May 11 at 23:25
• I misread; sorry. How about: if $d(x,F)=0$ then $x\in F$ contrary to hypothesis. Otherwise, $d(e,F)\ge f(x,F)>0$ for all $e$, so $d(E.F)>0$. – kimchi lover May 11 at 23:37
• Ahh, right. I mean I wanted to get $d(E,F)$ in there somewhere, that's what was messing with me. But I couldn't say that $d(E,F) \geq d(x,F) > 0$,, but what you did was sort of like that, you said $d(e,F) \geq d(x,F) > 0$ for all $e$, which is kind of the same thing... There is some small technical detail that I feel like I'm not understanding. – Mathematical Mushroom May 11 at 23:38

$$F$$ 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.

• Right, $p$ is in the complement of $F$, a closed set, so we can find an $\epsilon$ neighborhood. Thanks man! – Mathematical Mushroom May 12 at 14:18
• So it's possible for $E$ to be compact, $E \cap F = \emptyset$, and then the minimum of our function at point $p$ to be in $F$ if $F$ is not closed???? But the sets are still disjoint, I'm confused how $p$ could ever be in $F$ whether $F$ is closed or not. – Mathematical Mushroom May 19 at 13:08
• @MathematicalMushroom The function $f$ is defined on $E$ so it has a minimum $p$ in $E$, by compactness. And of course $p \notin F$ as $E \cap F = \emptyset$. So then we can use that $F$ is closed in the final step. – Henno Brandsma May 19 at 13:11
• Right, to show the distance is greater than zero, not to conclude that $p \notin F$. Thank you for your patience. – Mathematical Mushroom May 19 at 13:13

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 $$d(E,F)=\inf \{d(e,F): e \in E\}=f(x)=d(x,F) >0$$.

• Thank you sir. However, I was trying to prove it without passing to a subsequence, but rather exploiting the compactness of our set by appealing to the extreme value theorem. Do you know how to do the proof this way? – Mathematical Mushroom May 11 at 23:39
• Also, in your comment, do you mean $(f_n)$ tends to $e$ rather than $(f_n_k)$? – Mathematical Mushroom May 11 at 23:40
• @MathematicalMushroom No we cannot show that $(f_n)$ tends to $e$. We only know that $(e_{n_k})$ tends to $e$. – Kavi Rama Murthy May 11 at 23:42
• Right, I'm sorry, i misread. You're answer makes me realize that I never used the fact that $F$ was closed in doing the proof the way I'm trying to do it. – Mathematical Mushroom May 11 at 23:43
• @MathematicalMushroom Your argument also works with just one more observation. I have edited my answer. – Kavi Rama Murthy May 11 at 23:47