Problem of the Week #100 - April 28th, 2014

[Chris L T521]

This is the 100th week we've had Graduate POTW problems on MHB! (Party)

When we started the POTW over two years ago, we weren't sure whether or not it would be ideal to have POTWs covering graduate level topics since we didn't have that many advanced members back then. However, nine weeks later, I decided to give things a try and it's been a somewhat bumpy road with this since then, but I'm glad that I've stuck with it. At this time, I'd like to extend my thanks to those of you have participated in the graduate POTWs every now and then; there have been many weeks when no solutions have been submitted, but then there are weeks when I get an few, so it's hit or miss...but I've loved every minute of it! (Smile)

Anyways, let's get back to our regularly scheduled program.

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Here's this week's problem!

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Problem: Let $(K,d)$ be a compact metric space and let $f:K\rightarrow K$ be a map such that $d(f(x),f(y))<d(x,y)$ for all $x\neq y$. Denote $K_0=K$ and define recursively $K_{i+1}=f(K_i)$. Prove that $\bigcap_{i=0}^{\infty} K_i$ is a one-point set.Hint: [sp]Let $A=\bigcap_{i=0}^{\infty}K_i$. Show that $A$ is compact and $f(A)=A$, then conclude that $A$ is a one-point set.[/sp]

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by girdav and Opalg. You can find Opalg's solution below.

[sp]First, notice that the condition $d(f(x),f(y)) < d(x,y)$ implies that $f$ is continuous and therefore takes compact sets to compact sets. Next, $K_1 = f(K) \subseteq K = K_0$. Therefore $K_2 = f(K_1) \subseteq f(K_0) = K_1$, and similarly (by a simple induction argument) $K_{n+1} \subseteq K_n$ for all $n$. Thus the sets $K_n$ form a decreasing nested sequence of nonempty compact sets. By Cantor's intersection theorem their intersection $A$ is also a nonempty compact set.

If $x\in A$ then $x\in K_n$ for all $n$ and so $f(x)\in K_{n+1}$ for all $n$. Therefore $f(x)\in A$. So $f(A) \subseteq A$. Also, for each $n\geqslant 2$, $x\in K_n = f(K_{n-1})$ so that there exists $x_n\in K_{n-1}$ with $f(x_n) = x$. By compactness, the sequence $(x_n)$ has a convergent subsequence $x_{n_k}\to y \in K$. But $x_m \in K_n$ whenever $m > n$. So for any given $n$, $x_{n_k}\in K_n$ for all sufficiently large $k$. But $K_n$ is closed and therefore $y\in K_n$. Since that holds for all $n$, $y \in A$. By continuity of $f$, $f(x_{n_k}) \to f(y)$ as $n\to\infty.$ But $f(x_{n_k}) = x$ for all $k$, and so $x = f(y)$. Therefore $A\subseteq f(A).$ Thus $f$ maps $A$ onto itself.

Now suppose that $A$ contains more than one point. The map $(x,y) \mapsto d(x,y): A\times A \to\mathbb{R}$ is a continuous function on a compact set and therefore attains its maximum value, $D$ say. Thus there exist distinct points $u,v \in A$ with $d(u,v) = D.$ Since $f:A\to A$ is surjective, there exist distinct points $s,t \in A$ with $f(s) = u$ and $f(t) = v.$ But then $D = d(u,v) = d(f(s),f(t)) < d(s,t)$, which contradicts the maximality of $D$. Therefore $A$ cannot contain more than one point, and so $A = \bigcap_{n=0}^\infty K_n$ is a one-point set.[/sp]