(June 2013 #5b) Let \((X,d)\) be a complete metric space, \(A\subset X\) be a bounded set, and \(F:X\to X\). Assume there exists some \(k>0\) such that \[d(F(a),F(b))\le kd(a,b)\qquad\text{for all }a,b\in X\] and that \(A\subset F(A)\). Provide a description of \(A\) if \(k<1\). Can anything be said about \(A\) if \(k\ge1\)?
Solution
1. Case \(k < 1\): \(A\) is a singleton.Proof: \(A \subseteq F(A)\) gives \(\mathrm{diam}(A) \le \mathrm{diam}(F(A))\) (a subset has no larger diameter).Since \(F\) is \(k\)-Lipschitz with \(k < 1\), \(\mathrm{diam}(F(A)) \le k\,\mathrm{diam}(A)\).Hence \(\mathrm{diam}(A) \le k\,\mathrm{diam}(A)\), forcing \(\mathrm{diam}(A) = 0\) (as \(k < 1\) and \(\mathrm{diam}(A) < \infty\) by boundedness).So \(A\) contains at most one point, i.e. \(A\) is either empty or a singleton. If \(A = \{a\}\) is a singleton, then \(A \subseteq F(A) = \{F(a)\}\) forces \(F(a) = a\), so \(A\) is the singleton containing the unique fixed point of the contraction \(F\).(Uniqueness: two fixed points \(a \ne a'\) would satisfy \(d(a,a') = d(F(a),F(a')) \le k\,d(a,a') < d(a,a')\), impossible.)Existence of the fixed point is the Banach contraction theorem (complete space, \(k < 1\)), and \(A\) must be exactly that point.
2. Case \(k \ge 1\): nothing can be said in general.Proof: take \(X = \mathbb{R}\) with the usual metric (complete) and \(F = \mathrm{id}\), \(k = 1\).Then \(A \subseteq F(A) = A\) holds for every bounded nonempty \(A\), and \(A\) can be any bounded set (a point, an interval, infinitely many points).So no structural description of \(A\) is possible from the hypotheses alone when \(k \ge 1\).
3. Q.E.D.