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In trying to see how Banach's fixed point theorem would break down in an incomplete space, I tried to come up with an example of a function: $f: \mathbb{Q} \longrightarrow \mathbb{Q} \ \ $such that $ \forall x,y \in \mathbb{Q}$ our function is contracted with factor $\frac12$ and has no fixed points. i.e. $$ \left | {f(x)-f(y)} \right |\leq \frac12 \left | x-y \right | ,\\ \nexists \ \ x^* \in \mathbb{Q} \ \ s.t. \ f(x^*)=x^* \ $$ Starting with $f(x)=x^2-2$ and using Newton-Raphson I was able to construct the following function which works $$ f(x) := \left\{\begin{array}{lr} \frac12 (x-1) + \frac32, & \text{for } x < 1\\ \frac12- x^{-2}, & \text{for } 1\leq x\leq 2\\ \frac12 (x-2) + \frac32, & \text{for } x>2 \end{array}\right\}$$

That led me to thinking if there was a bijection which satisfied the same criteria. I think I can see why one could exist but am not sure how to prove that one necessarily exists. I am also keen on an explicit example of a bijection that works.

Gridley Quayle
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  • Can't you just pick anything that has an attractive irrational fixed point? e.g., a fixed point of the function $f$ defined by $f(x)=e^{-x}$ is the irrational Omega constant $\omega\approx 0.567\ldots$ (https://en.wikipedia.org/wiki/Omega_constant) and $|f^\prime|<1$ on $(0,\infty)$. – parsiad Jan 24 '17 at 18:38
  • @parsiad I don't think so because the image of our function has to be a subset of the rationals. – Gridley Quayle Jan 24 '17 at 18:42

2 Answers2

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Consider for example $$ f(x) = \begin{cases} x/2 + 3/2 & x\le 2 \\ 3-1/x & 2 \le x \le 4 \\ x/2 + 3/4 & x \ge 4 \end{cases} $$ In the reals the only fixpoint is at $x=\frac{3+\sqrt5}2 \approx 2.6 $, which is irrational.

Because the curved segment in the middle uses $x^{-1}$ instead of $x^{-2}$, its inverse is a rational function, so the full $f$ is a contractive bijection $\mathbb Q\to\mathbb Q$.

hmakholm left over Monica
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A much simpler example is $f : \mathbb R \backslash \mathbb Q \to \mathbb R \backslash \mathbb Q$ defined by $$f(x)=\frac{x}{2}$$

This is obviously a contraction, but there is no fixed point since $0 \notin \mathbb R \backslash \mathbb Q$.

N. S.
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