Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $T$.

share|cite|improve this question
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

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 $d(x, y) \ge d(x, T) \ge d(x_0, T) = d(x_0, y_0)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.