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Let $S\subset\Bbb R$ not empty, define $f:\Bbb R\rightarrow\Bbb R$ such that $f(x)= \inf\{|x-s| ;s\in S\}$

then, prove that $|f(x)-f(y)|\le|x-y| $ for any $x,y \in \Bbb R$

  • 1
    Have you tried the case where $S$ consists of a single point?2012-09-08

1 Answers 1

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$|x - s| \le |x - y| + |y - s|$ for every $s \in S$. Hence $f(x) \le |x - y| + f(y)$.

Similarly $f(y) \le |x - y| + f(x)$.

Hence $|f(x) - f(y)| \le |x - y|$.

  • 1
    Indeed, the distance function on any metric space is continuous, by the same proof that you have given.2012-09-08