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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!


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Why do you think this should be simple? – Chris Eagle Aug 28 '12 at 12:49
Presumably $S$ and $T$ have to be non-empty so $x$ can be on the boundary of $S$ and $y$ on the boundary of $T$. – Henry Aug 28 '12 at 12:52
Chris Eagle: Well, I guess it is simple because it is used in a bigger proof without being sub-proved. – Tanja Aug 28 '12 at 12:56
Henry: Indeed, $S$ and $T$ can be assumed to be non-empty. – Tanja Aug 28 '12 at 12:56
up vote 1 down vote accepted

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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Looks like a helpful answer! The set $d^{c}$ is bounded by definition, and it is intuitively closed. So I can use the Weierstrass extreme value theorem. The function $d: S \times T \longrightarrow \mathbb{R}$ only needs to be continuous for that, not coercive, right? Thank you anyway for this good answer. – Tanja Aug 28 '12 at 13:16
Be careful: $d^c$ is trivially closed, but it is not trivially bounded. Imagine the case where $S$ is a hyperbola and $T$ is one asymptote. There are points of the two sets that lie close and yet they "diverge" to infinity. Actually, the distance between this $S$ and this $T$ is zero, and it cannot be attained. – Siminore Aug 28 '12 at 13:19
You are right, in principle one could construct cases where $d^{c}$ isn't bounded. It is not bounded "by definition". Yet, given $S$ being bounded, my intuition tells me that this is not possible (a hyperbola which is "contained in a large ball" couldn't have an asymptote). What is your simplest argument that $d^{c}$ is closed? (Sorry, I am am a bit lazy today.) – Tanja Aug 28 '12 at 13:30
@Tanja the function $d \colon S \times T \to \mathbb{R}$ is continuous, so that $d^c = d^{-1}([0,c])$, and thus closed. – Siminore Aug 28 '12 at 13:41
The approach is probably possible, but the version of the Weierstrass extreme value theorem which I know works only for functions $S \longrightarrow \mathbb{R}$, where $S \subseteq \mathbb{R}^{n}$. Here, however, the Euclidean distance function $d$ maps $d^{c}$ into $\mathbb{R}$, where $d^{c} \subseteq \mathbb{R}^{n \times n}$. Possibly the Weierstrass theorem can be generalized, but for my purposes that is too much effort. I need to come up with a simpler proof. – Tanja Aug 28 '12 at 13:41

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