Let $E$ and $F$ be a compact set on un metric space $(X,d)$. Show that dist$(E,F)>0$ if and only if $E \cap F = \emptyset $

Question

Let $E$ and $F$ be a compact set on un metric space $(X,d)$. Show that dist$(E,F)>0$ if and only if $E \cap F = \emptyset $

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($\Leftarrow$)

if $E \cap F = \emptyset \ \ \ \ \ \text{then} \ \ \ \ \forall x,y ,\ \ \ \ \ x \in E, \ \ \ \ y \in F $ then $$inf \{ d(x,y)|x \in E , y \in F \}>0 \implies \text{dist}(E,F)>0$$

($\Rightarrow$)
my idea is to suppose that $E \cap F \ne \emptyset.$ Let $z \in E\cap F$. Suppose that $\{x_n\}$ and $\{y_n\}$ both converge to z because the set E and F are compact. Then arrive at a contradiction that dist$(E,F)=0$

Is my proof correct?

Answer 1

Actually, the second part is trivial, you don't need the compactness. The real place you need the compactness is in the first part. Prove d is continuous, since E and F are compact, d can attain its minimum on $E\times F$, then the inf is positive.

Answer 2

Since E is compact it is sequentially compact. Chose an abritrary $\{ xn \} \subseteq E $ which the subsequence $\{x_{n_k}\} \rightarrow x_0$ in E and $\{ yn \} \subseteq F $ with $\{y_{n_l}\} \rightarrow y_0$ in F. Then $d(x_n,y_n)>0 $. Since $x_0 \in E$ and $y_0 \in Y $ then d$(x_0,y_0)>0$.

Since $E \cap F = \emptyset$ , It is also true that $d(x_n, y_n) >0$. Then we get that:

$$inf \{ d(x,y,) | x \in E , y \in F \} > 0 $$

Answer 3

If inf=0, then there are sequences $\{x_n\}$ and $\{y_n\}$ such that d(xn,yn) converges to 0. Pick subsequences xnk and ynk converges to x0 and y0 in E and F respectively. Then d(x0,y0)=0, which is a contradiction. Is that what you mean?