If $F$ and $K$ are closed sets and $d(F,K)=0$ does it follow that $F\cap K\neq \emptyset$? I can't find an example that it doesn't follow but I can't prove it either.
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Hint: $(-1,0)\cup(0,1)$ (there are other examples)
Another: $F=\{(x,y)\in\mathbb{R}^2:y=0\}$, $K=\{(x,y)\in\mathbb{R^2}:x\ne0, y=1/x\}$
Following the discussion in comments, if $K$ is compact then the conclusion is true. Indeed, if $d(F,K)=0$, there exist sequences $(x_n)$ in $F$ and $(y_n)$ in $K$ such that $d(x_n,y_n)\to0$. Since $K$ is compact, a subsequence $(y_{n_k})$ of $(y_n)$ is convergent. Since $$ \lim_{k\to\infty}d(x_{n_k},y)=0 $$ (prove it) we can conclude that $(x_{n_k})$ converges to $y$. Since $F$ is closed, $y\in F\cap K$.
egreg
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but is the first one closed? – TanEma Aug 13 '17 at 13:42
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@TanEma The first is the whole space; find $F$ and $K$ yourself. – egreg Aug 13 '17 at 13:43
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@TanEma: Note that in egreg's second example the sets are closed, so indeed we cannot prove a nonempty intersection in general. However if one of the sets $F,K$ is compact, then you can prove it for a complete metric space. – hardmath Aug 13 '17 at 13:44
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@hardmath why do you need completeness? – Jens Renders Aug 13 '17 at 14:55
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@JensRenders: It is redundant; I mentioned it for contrast with the first example. – hardmath Aug 13 '17 at 15:28
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Not necessarily. For instance, take: \begin{align*} F := \{(x,y) \in \mathbb{R}^2 : y \geq \exp(x) \}\\ K := \{(x,y) \in \mathbb{R}^2 : y \leq -\exp(x)\}\\ \end{align*}
which are both closed as inverse image of closed sets by continuous applications. As $x \rightarrow -\infty$, the two sets get closer and closer to each other, but they never intersect.
To make this counterexample, we had somehow to use the fact that both $F$ and $K$ were not bounded. We can show that if either $K$ or $F$ is compact, then necessarily $F \cap K \neq \emptyset$.
Hermès
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