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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 $zin X$, 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, $(forall epsilon >0) (exists K in mathbbN)$ 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.

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 $m>ngeq 1$ 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 $F^n_k(z)to ain X$.

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 $epsilon$.

There exists a subsequence $(F^n_k(z))_k$ which is a Cauchy sequence. Thus, for $varepsilon > 0$ there exists $K$ such that for $kgeellge K$ 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.