Introductory Example for Fixpoints

Let $A = [0,1] = \{ x \in \mathbb{R} \mid 0 \le x \le 1 \}$ be the interval of real numbers. Recall that, by definition of real numbers and complete lattice, $(A,\le)$ is a complete lattice with least lattice element $0$ and greatest lattice element $1$ . Here $\sqcup$ is the least upper bound operator on sets of real numbers, also called supremum and denoted sup in real analysis.

Let function $f : A \to A$ be given by

$\begin{equation*} f(x) = \left\{\begin{array}{l} \frac{1}{2} + \frac{1}{4}x, \mbox{ if } x \in [0,\frac{2}{3}) \\ \ \\ \frac{3}{5} + \frac{1}{5}x, \mbox{ if } x \in [\frac{2}{3},1] \end{array}\right. \end{equation*}$

(It may help you to try to draw $f$ .)

Part a)

Prove that $f$ is monotonic and injective (so it is strictly monotonic).

Part b)

Compute the set of fixpoints of $f$ .

Part c)

Define $iter(x) = \sqcup \{ f^n(x) \mid n \in \{0,1,2,\ldots \}\}$ . (This is in fact equal to $\lim_{n\to\infty} f^n(x)$ when $f$ is a monotonic bounded function.)

Compute $iter(0)$ (prove that the computed value is correct by definition of $iter$ , that is, that the value is indeed $\sqcup$ of the set of values). Is $iter(0)$ a fixpoint of $f$ ? Is $iter(iter(0))$ a fixpoint of $f$ ?