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I would like to prove the following statement: Let $S,T \subseteq \mathbb{R}^{n}$ be closed sets with $S \cap T = \emptyset$, at least one of which is bounded. Then there exist $x \in S$ and $y \in T$ such that $d(x,y) \leq d(\hat{x},\hat{y}) \text{ for all } \hat{x} \in S, \hat{y} \in T,$ where $d(\cdot,\cdot)$ is the Euclidean distance. Should be simple, but couldn't find the proof immediately. Could anyone help me please? Thanks a lot!

Tanja

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    Henry: Indeed, $S$ and $T$ can be assumed to be non-empty.2012-08-28

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Assume $S$ is bounded. For every $c>0$, consider the set $ d^c = \left\{ (x,y) \in S \times T \mid d(x,y) \leq c \right\}. $ Try to check that $d^c$ is closed and bounded: it is rather clear, since $S$ is contained in a large ball and thus there cannot exist points in $T$ that lie both close to $S$ and arbitrarily far from the origin. Then you want to minimize a continuous and coercive function, a standard generalization of Weierstrass' theorem.

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    If $\{x_n\}$ is bounded and $\{d(x_n,y_n)\}$ is bounded, how can $\{y_n\}$ be unbounded? For example, $|d(y_n,0)-d(x_n,0)| \leq d(x_n,y_n)$, and therefore $d(y_n,0) \leq \ldots$2012-08-28