3

Let $S \subset \mathbb R$ be any set, and define for any $x \in \mathbb R$ the distance between $x$ and the set $S$ by $d(x,S) = \inf\{|x-s| : s \in S\}$.

  1. Prove that the function $d_s: \mathbb R \to [0,+\infty)$ given by $d_s(x) = d(x,S)$, is Lipschitz continuous.

  2. Prove that if $S$ is compact then for every $x$ in $\mathbb R$, there is $s$ in $S$ such that $|x-s| = d(x,S).$

Caleb Stanford
  • 45,895
Tilly
  • 51
  • 2
    This question is about to receive a serious amount of down votes and even may be closed. You must include some work you've tried and learn to use latex formatting. – Faraad Armwood Aug 16 '16 at 20:17
  • Thanks for the help, first time using this website but i wouldn't have posted if i knew where to start. I understand the definition of lipschitz however i guess I'm stuck understanding the initial statement. I would have loved to use latex on this website but i did not see where it would let me. I was typing as i was using latex but it did not recognize any of the commands. – Tilly Aug 16 '16 at 20:20
  • Okay, I will edit the question and give you some hints. – Faraad Armwood Aug 16 '16 at 20:22
  • Thank you, i appreciate it. – Tilly Aug 16 '16 at 20:24
  • For #1, consider the following slightly simpler situation to get the main idea. You fix a point $x$ and let's assume that the the minimum distance to $S$ exists and is achieved by $s.$ Now wiggle $x$ around a bit by an amount $\epsilon.$ Then the distance of the wiggled $x$'s from $S$ can't be more than $\epsilon$ since that's how far the wiggled $x$'s will be from $s$ (indeed, there might be elements in $S$ that are closer to the wiggled $x$'s than $s$ is). – Dave L. Renfro Aug 16 '16 at 20:35
  • Regarding my too hastily written comment, the distance of the wiggled $x$'s to $S$ can't differ from the distance of $x$ to $S$ by more than $\epsilon$ since . . . – Dave L. Renfro Aug 16 '16 at 20:57
  • The first question is answered here – Caleb Stanford Aug 16 '16 at 21:13
  • @6005 how does that question relate to the Lipschitz continuous definition since if a function is continuous it is not necessarily Lipschitz continuous – Tilly Aug 16 '16 at 22:00
  • @Tilly If you read the question it says to prove that $|f(x) - f(y)| \le |x - y|$. – Caleb Stanford Aug 16 '16 at 22:02
  • Ahh yea i missed that! Thank you. – Tilly Aug 16 '16 at 22:07

1 Answers1

1

Hints:

For the first part, observe that $d(x,s)\leq d(x,y)+d(y,s)$ and $d(y,s)\leq d(x,y)+d(x,s)$.

For the second part, note that for fixed $x,\ $ $f(s)=\vert x-s\vert $ is continuous on the compact set $S$.

Matematleta
  • 29,139
  • For the first part, i guess i don't see how you came up with that $d(x,s)$ $\ge$ $d(x,y) + d(y,s). $The definition that i have includes K being the constant such that K $\ge$ 0$ @Chilango – Tilly Aug 16 '16 at 20:56
  • I think you have the inequality reversed. Anyway, in my answer, it's just the triangle inequality where $d$ is absolute value, and $s$ is any element of $S$. You will use this before passing to the infimum. – Matematleta Aug 16 '16 at 21:01
  • Okay i understand the first part using the triangle ineq. But what do you mean before passing the infimum, since we are proving that $d_s$ :$R$ $\to [o,+inf)$, $d_s$(x)= d(x,S) is L.C? @chilango – Tilly Aug 16 '16 at 21:37