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let $S$ and $T$ be two disjoint compact nonempty sets. Show that there are points $x_0$ in $S$ and a point $y_0$ in $T$ such that $|x-y| \geq|x_0 -y_0|$ whenever $x$ is in $S$ and $y$ is in $T$.

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You should say in which space you're working. But since you use $|x-y|$ to denote the distance, I guess $S,T\subseteq\mathbb R$. –  Martin Sleziak Jun 1 '12 at 10:08
    
It might be useful to start by showing that the function $x\mapsto d(x,S)$ is continuous, see also here. The distance between a point $x$ and a set $S$ is defined by $d(x,S)=\inf_{s\in S} |x-s|$. –  Martin Sleziak Jun 1 '12 at 10:15

1 Answer 1

Since you use the notion of distance, $S$ and $T$ should be subsets of a metric space $X$. Let us denote the distance in $X$ by $d$.

Define $d(x, T) = \inf_{y \in T} d(x, y)$ for $x \in S$. Prove that $x \mapsto d(x, T)$ is a continuous function. Then it reaches it's minimum, since $S$ is compact. Thus there is $x_0 \in S$ such that $d(x_0, T) \le d(x, T)$ for each $x \in S$. Now consider the function $y \mapsto d(x_0, y)$. Prove that it is continuous. Since $T$ is compact, it reaches it's minimum. So there is $y_0$ such that $d(x_0, y_0) \le d(x_0, y)$ for each $y \in T$.

Now let $x \in S, y \in Y$, then $d(x, y) \ge d(x, T) \ge d(x_0, T) = d(x_0, y_0)$.

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