7
$\begingroup$

Suppose $(X,d)$ is a compact metric space and $f:X\to X$ a continuous map. Show that $f (A)=A$ for some nonempty $A\subseteq X.$

I start this by supposing that $A_0:=X$ and $A_{n+1}:=f(A_n)$ for all $n \geq 0$. If $A_n=A_ {n+1}$ for some $n$ then the purpose is done. But if not, how can we think further?

$\endgroup$
  • $\begingroup$ What is the question? $\endgroup$ – Pedro Tamaroff Apr 9 '13 at 16:35
  • $\begingroup$ Sorry but question as it stands does not make any sense. Are you trying to say something about the limiting behaviour of $A_n$? Are you trying to approximate an attractor of some sort? $\endgroup$ – muzzlator Apr 9 '13 at 16:35
  • 1
    $\begingroup$ @muzzlator "Show that $f(A)=A$ for some nonempty $A$ subset of $X$" makes sense for me. $\endgroup$ – Julien Apr 9 '13 at 16:43
  • $\begingroup$ @julien It makes sense to me too, I don't know how but I must have completely missed that line when I first read it $\endgroup$ – muzzlator Apr 9 '13 at 16:45
  • 1
    $\begingroup$ @muzzlator: It got added after the initial post, but before the initial $5$-minute free-editing period expired. $\endgroup$ – Cameron Buie Apr 9 '13 at 16:47
4
$\begingroup$

First, show that each of your $A_n$ is closed in $X$. It is also useful to note that $A_{n+1}\subseteq A_n$ for all $n$.

Then let $A=\cap_{i=1}^n A_n$ using $A_n$ from your definition.

Show that $f(A)=A$.

Finally, if $A$ is empty, show that $\cup_n (X\setminus A_n)$ is an open cover of $X$. Can we find a finite sub-cover of this cover?

(Alternatively, you could also use the sequence definition for compactness in metric spaces. Pick any $x_0\in X= A_0$. Define $x_{n+1}=f(x_n)$. Then $x_n\in A_n$. The sequence must have a convergent subsequence - show that limit of that subsequence is in $A_n$ for all $n$.)

$\endgroup$
  • $\begingroup$ Shouldn't you have something like $A= \bigcap _{i = 1}^\infty A_i$ and not $A_n$? $\endgroup$ – Stefan Apr 9 '13 at 16:57
  • 1
    $\begingroup$ Why $A_{n+1}\subseteq A_n$? $\endgroup$ – Julien Apr 9 '13 at 16:58
  • 1
    $\begingroup$ @julien Induction. Or directly, $A_{n+1}=f^{n+1}(X)=f^{n}(f(X))\subseteq f^{n}(X)=A_n$. $\endgroup$ – Thomas Andrews Apr 9 '13 at 16:59
  • 1
    $\begingroup$ Oh boy...silly me. $\endgroup$ – Julien Apr 9 '13 at 17:01
  • 1
    $\begingroup$ Ah, just for completeness, the "f.i.p" is the "finite intersection property." A family of subsets satisfies the FIP if any finite set of elements of the set has non-empty subset. There is a dual to the "open cover" property of compactness in terms of collections of closed subsets that satisfy the FIP. $\endgroup$ – Thomas Andrews Apr 9 '13 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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