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please how to solve this question

Let $E$ a compact metric space and a function $f:E\to E$ such that $$ \forall x,y \in E, d(f(x),f(y))\geq d(x,y)$$ Let $a\in E$ and a sequence $(f^n(a))$ , prove that $a$ in an adherent value of $(f^n(a))$.

As E is compact then $(f^n(a))$ has a convergent subsequence, but I stop here

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Hint: Prove by contradiction. Suppose $(f^n(a))$ stays away from $a$, so there is some $\varepsilon>0$ such that $d(f^n(a),a)>\varepsilon$ for all $n\geq 1$. Prove that $(f^n(a))_{n>m}$ also stays away from $f^m(a)$ for all $m$ (with the same $\varepsilon$), and show this yields a contradiction with compactness of $E$.

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  • $\begingroup$ why we suppose that the hole sequence do not converge to a? we just need a subsequence $\endgroup$ – Poline Sandra Jun 12 at 15:59
  • $\begingroup$ That's why you need the whole sequence avoids being arbitrarily near $a$, so there is no way a subsequence converges to $a$. $\endgroup$ – user10354138 Jun 12 at 16:01
  • $\begingroup$ so we prove by contradiction $\endgroup$ – Poline Sandra Jun 12 at 16:05
  • $\begingroup$ read first sentence. $\endgroup$ – user10354138 Jun 12 at 16:14
  • $\begingroup$ I suppose that there is no subsequence which converge to a, but i don't understand how to use the condition of f ? $\endgroup$ – Poline Sandra Jun 12 at 16:16

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