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If $X$ is a compact metric space, $A: X\to X$, is it true that if $a = \inf d(x,Ax),\space x \in X$, then there exists $y \in X$ such that $d(y,Ay) = \inf d(x,Ax)$? If so, why?

Thomas Russell
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Simon Sehayek
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No, it's not true. Let $X = [0,1]$. Define $A$ by $Ax = 1$ for $x\in [0,1)$, and $A1 = 0$. Then $a=0$ but always $d(x,Ax) > 0$. (Obviously, the statement is true if $A$ is continuous.)

Yury
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  • oh wait... I get it – Simon Sehayek Oct 14 '12 at 22:33
  • can I refer you to this proof... http://math.stackexchange.com/questions/118536/prove-the-map-has-a-fixed-point. The answer uses the fact that K is compact to say that this holds... – Simon Sehayek Oct 14 '12 at 22:38
  • In that question $A$ is a contraction, therefore, it is a continuous function. – Yury Oct 14 '12 at 22:43
  • if A is a contraction, wouldn't it follow immediately that there exists a unique fixed point – Simon Sehayek Oct 14 '12 at 22:45
  • It depends on your definition of “immediately” :-). Yes, it is a true that every contraction $A$ has a unique fixed point, if $X$ is a compact metric space (or if $X$ is a complete metric space). – Yury Oct 14 '12 at 22:56
  • Actually, in that question, the mapping $A$ decreases distances, $d(Ax, Ay) < d(x, y)$ but it is not apriori a contraction. Still $A$ must be continuous (so for such $A$ the statement from your question holds). – Yury Oct 14 '12 at 23:00
  • I don't see how it is so obvious that if A is continuous my assumption would make my question true. Can you just quickly outline a proof? – Simon Sehayek Oct 14 '12 at 23:03
  • No, you need to assume that $X$ is compact and $A$ is continuous. If $A$ is a contraction, then it is enough to assume that $A$ is complete. However, your statement is not true if you just assume that $X$ is complete. Consider the map $f\colon [1,\infty) \to [1,\infty)$ defined by $f(x) = x+ 1/x$. For it, $a=0$ but $f(x)\neq x$. – Yury Oct 14 '12 at 23:08
  • Right so under those assumptions... how would I show my question holds – Simon Sehayek Oct 14 '12 at 23:10
  • If $A$ is continuous and $X$ is compact, just consider a sequence $x_n$ such that $d(x_n, Ax_n) < a + 1/n$ then find a limit point of $x_n$. – Yury Oct 14 '12 at 23:10
  • Ahh cool, the sequence is monotone decreasing and has limit a. Thanks for your help! – Simon Sehayek Oct 14 '12 at 23:16