Let $(X,d)$ be a metric space. Let $E$, $F$ be two disjoint non-empty subsets of $X$ with $E$ compact and $F$ closed.
Show that $\inf\{d(x,y): x\in E, y\in F\}>0$
Show that this does not longer true is $E$ is not compact: find two disjoint closed subsets $E$ and $F$ of $\mathbb{R}^2$ so that $\inf\{d(x,y): x\in E, y\in F\}=0$
I have been trying to use the fact sequential compactness iff compact on a metric space. I haven't had much luck. Any help would be hugely appreciated.
If you want to use sequential compactness, you can argue as follows: Suppose $\inf\{d(x,y) \mid x\in E, y\in F\} = 0$. Then there are $x_n \in E$, $y_n \in F$ with $d(x_n, y_n) \to 0$. As $E$ is compact, $(x_n)$ has a convergent subsequence, say $x_{n_k} \to x \in E$. Then $d(x, y_{n_k}) \to 0$. As $F$ is closed, $x\in F$, contradiction to $E \cap F = \emptyset$. For the second question think of sets $E$ and $F$ touching "at infinity".
Hint. Can you show that the function $f : E \to \mathbb{R}$ defined as $f(x) = \inf \{d(x, y) : y \in F\}$ is continuous and greater than $0$?
