Hint or help on this metric spaces problem?

Question

Let $(X,d)$ be a sequentially compact metric space. Let $F$ map $X$ into $X$, with the property $d(F(x),F(y)) \geq d(x,y)$ for all $x$ and $y$ in $X$.

Show that every point $z$ in $X$ is a cluster point (i.e, in the closure) of the set $\{F^{n} (z) : n = 1,2,...\}$

I'm just not entirely sure how to use the fact that the distance of the images is at least the distance of the preimage.

What I have so far:

Let $\epsilon > 0$. Consider an open ball around $z\in X$, with radius epsilon; call it $B(z;\epsilon)$. Then since $X$ is sequentially compact we may form a converging subsequence from any sequence in $X$. Hence, $(\forall \epsilon >0) (\exists K \in \mathbb{N})$ so that $z_{n_{k}} \in B(z;\epsilon) \forall k>K$

The idea here is to try to show that the intersection of this ball and the above set is nonempty, hence showing that $z$ is in its closure.

Is this the wrong way to start? Am I missing something? Again, it isn't too clear to me how I'm to bring in the other assumption.

Answer 1

I will take the 'in the closure' as your definition of 'cluster point' since, for example, for $X$ the closed unit disc with $F$ a rotation, say of angle $\pi$, satisfies $d(F(x),F(y))=d(x,y)$, but then $\{F^n(0)\}=\{0\}$ doesn't accumulate at $0$ in the sense of every neighborhood having elements of $\{F^n(0)\}$ different from $0$.

If there is $n>0$ such that $F^n(z)=z$ we are done since the its orbit is finite and therefore closed, and it contains $z$.

Let's assume for the rest that $f^{n}(z)\neq z$ for $n>0$.

It can't happen that the orbit is eventually periodic. If it were so, assume that $m>n\geq 1$ are smallest such that $f^m(z)=f^n(z)$, then $0=d(f^m(z),f^n(z))\geq d(f^{m-n}(z),z)>0$ is a contradiction.

So, we can assume that $\{F^n(z)\}$ is infinite. Since $X$ is compact this sequence has a convergent subsequence $F^{n_k}(z)\to a\in X$.

Take $\epsilon >0$, then there is $K$ such that for $k>K$ we have $\epsilon>d(F^{n_{k+1}}(z),F^{n_k}(z))\geq d(F^{n_{k+1}-n_k}(z),z)$. Therefore $F^{n_{k+1}-n_k}(z)$ is inside the ball with center $z$ and radius $\epsilon$.

Answer 2

There exists a subsequence $(F^{n_k}(z))_k$ which is a Cauchy sequence. Thus, for $\varepsilon > 0$ there exists $K$ such that for $k\ge\ell\ge K$ we have $d(F^{n_k}(z),F^{n_\ell}(z)) < \varepsilon$. Hence, by induction, $d(F^{n_k-n_\ell}(z),z) < \varepsilon$. Since this holds for any $\varepsilon > 0$, that's it.