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Compactness, limit point and subsequence (from an analysis test)

Dense

New member
May 16, 2012
1
Let $(E,d)$ be a compact metric space. Let $f:E\to E$ be a function such that for all $x,y\in E$ is $d(f(x),f(y))\ge d(x,y).$

  1. Let $a\in E.$ Prove that $a$ is a limit point of the sequence $(f^n(a))_{n>0}.$ ($f^n(a)$ means the composition of $f,$ $n-$ times. Example: $f^3(a)=f(f(f(a))).$ )
  2. Conclude that $f(E)$ is dense on $E.$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
Let $(E,d)$ be a compact metric space. Let $f:E\to E$ be a function such that for all $x,y\in E$ is $d(f(x),f(y))\ge d(x,y).$
  1. Let $a\in E.$ Prove that $a$ is a limit point of the sequence $(f^n(a))_{n>0}.$
Proof by contradiction: suppose the result is false. Then there exists $\varepsilon>0$ such that $d(f^k(a),a) > \varepsilon$ for all $k\geqslant1.$ The condition on $f$ then implies that $d(f^{k+1}(a),f(a)) > \varepsilon$, and (by induction) $d(f^{k+l}(a),f^l(a)) > \varepsilon$ for all $l\geqslant0.$ In other words, $d(f^m(a),f^n(a)) > \varepsilon$ for all $m\ne n.$ Now get a contradiction by using compactness.