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I am aware the the question has been posted here:
For any subset $S$ of $\mathbb{R}$, the distance function $d_s(x) = d(x,S)$ is continuous
However, I would like feedback on my work.

Prove that the $\mathbb{R} \to [0,+\infty), d_s(x) = d(x,S)$, is Lipschitz continuous, provided that $s\subset \mathbb{R}$ be any set, and define for any $x\in \mathbb{R}$ the distance between $x$ and the set $S$ by $$d(x,S) = \inf\{|x-s| : s\in S\}$$

How I worked on it:
since Lipschitz continuous functions are by $|f(y) - f(x)| \leq K|y-x|$ where $K$ is a constant $\forall\, y \in S$ sufficiently near to $x$.

$d_s: \mathbb{R}\to [0,\infty) \implies d_s(x) = d(x,S)$
$|d_s(y)-d_s(x)| = |d(y,S) - d(x,S)|$
$\leq |y-x|$ by the property $|d(x,z)-d(y,z)| \leq d(x,y)$

$\leq 1\cdot|y-x|$

So $d_s(x)$ is Lipschitz continuous with $K= 1$.

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johnson
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    This was already asked here today: http://math.stackexchange.com/questions/1894378/for-any-subset-s-of-mathbbr-the-distance-function-d-sx-dx-s-is-c – Faraad Armwood Aug 17 '16 at 04:50
  • I want critique on my work as well though should i just post on that comment then? – johnson Aug 17 '16 at 05:04
  • When you notice that a similar (or exact) question has been posted, you have to try to make it very clear in the post that your question is not a duplicate. Otherwise, reviewers will try to close it as a duplicate of the previous post. They don't always read the posts carefully, that is why you have to try to make it clear. Also, don't write choppy formatting. For example, write $s\in S$,($s\in S$) not s$\in$ S (s$\in$ S). – Em. Aug 17 '16 at 05:58
  • ah thanks duly noted. – johnson Aug 17 '16 at 06:03

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