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Let $(X,d)$ be a compact metric space and let $T:X\longrightarrow X$ be a map such that $d(T(x),T(y))<d(x,y)$ for all $x,y\in X$ such that $x\neq y.$

(a) Prove that $T$ has a fixed point

(b) Prove that the fixed point is obtained as a limit of $\{T^{k}(x_{0})\}$ for $x_{0}\in X$

I have solved (a), however, I'm not able to prove (b).

In particular, I am having some problems using the compactness hypothesis on $X$ to find proper subsequences which are convergent.

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  • $\begingroup$ For part (b) it seems like you can use the fact that $d(T(x),t(y))<d(x,y)$ to construct a strictly decreasing sequence of positive numbers that is bounded below by zero. $\endgroup$ – graydad Oct 1 '14 at 15:12
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The following is probably not the most elegant solution, but nevertheless works.

Have a look at this nice answer https://math.stackexchange.com/a/409832/151552.

Observe that for an arbitrary subsequence $T^{n_k}(x_0)$, we have $T^{n_{k_\ell}}(x_0) \to x$ for some $x \in K$ and a suitable sub(-sub-)sequence (by compactness).

But $T^{n_{k_\ell}}(x_0) \in T^{n_{k_\ell}}(X) \subset \bigcap_{m=1}^N T^m (X) =: K_N$ as soon as $n_{k_\ell} \geq N$. Hence, $x \in \overline{K_N} = K_N$ (note that $K_N$ is closed).

As $N \in \Bbb{N}$ was arbitrary, $x \in \bigcap_N K_N = A$, where $A$ is chosen as in the answer linked above.

But the answer linked above shows that $A = \{\xi_0\}$, where $\xi_0$ is the unique fixed point of $T$. Hence, $x=\xi_0$.

Now use the fact that $(T^{n_k} x_0)_k$ was an arbitrary subsequence so conclude $T^n x_0 \to \xi_0$ (otherwise, there is $\varepsilon > 0$ and a subsequence $(T^{n_k}x_0)_k$ with $d(T^{n_k} x_0, \xi_0) \geq \varepsilon$ for all $k$, which contradicts the convergence $T^{n_{k_\ell}} x_0 \to \xi_0$ shown above).

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  • $\begingroup$ Thank you! Maybe it's possible to formulate just in terms of sub-sequences, but it works. So that I give you +1 $\endgroup$ – snaleimath Oct 2 '14 at 8:00
  • $\begingroup$ @PhoemueX Couldnt you say that $T^n(x_0) \rightarrow x$, by the iteration process, where $x$ is the fixed point, so for some $N > 0$ if $n > N$ you have that $d(T^n(x_0),x) < \varepsilon$, now just pick $jk > N$ and see that $d(T^{jk}(x_0),x) = d(T^N(T^{jk-N}(x_0)),x) < \varepsilon$ ? $\endgroup$ – Aram Oct 4 '14 at 7:09
  • $\begingroup$ I am sorry, I don't understand your question. Why do you mean with "by the iteration process"? If we already know that $T^n(x_0) \to x$, where $x$ is the/a fixed point, then we are done. $\endgroup$ – PhoemueX Oct 4 '14 at 14:48
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I would love to assume that there is no limit, then there would be at least two adherent values (due to compactness and continuity and infinity), then I may find contradiction somewhere.

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