How to show that a continuous map on a compact metric space must fix some non-empty set.

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?

• What is the question? – Pedro Tamaroff Apr 9 '13 at 16:35
• 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? – muzzlator Apr 9 '13 at 16:35
• @muzzlator "Show that $f(A)=A$ for some nonempty $A$ subset of $X$" makes sense for me. – Julien Apr 9 '13 at 16:43
• @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 – muzzlator Apr 9 '13 at 16:45
• @muzzlator: It got added after the initial post, but before the initial $5$-minute free-editing period expired. – Cameron Buie Apr 9 '13 at 16:47

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 $n$.)

• Shouldn't you have something like $A= \bigcap _{i = 1}^\infty A_i$ and not $A_n$? – Stefan Apr 9 '13 at 16:57
• Why $A_{n+1}\subseteq A_n$? – Julien Apr 9 '13 at 16:58
• @julien Induction. Or directly, $A_{n+1}=f^{n+1}(X)=f^{n}(f(X))\subseteq f^{n}(X)=A_n$. – Thomas Andrews Apr 9 '13 at 16:59
• Oh boy...silly me. – Julien Apr 9 '13 at 17:01
• 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. – Thomas Andrews Apr 9 '13 at 17:15