# Contraction function on $\mathbb{Q}$ that does not have a fixed point

We know the contraction principle holds on the complete metric space.

I am thinking about cases where the space is not complete, such as $$\mathbb{Q}$$. Could anyone help giving an example of a contraction function $$f: \mathbb{Q}\rightarrow\mathbb{Q}$$ that is ONTO and does not have a fixed point? How about if a bijection is required?

New contributor
Wei Liu is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.

Let $$t$$ be an irrational real number. Let rational approximations $$(a_n)_{n\geq 1}$$ strictly increasing to $$t$$ and $$(b_n)_{n\geq 1}$$ strictly decreasing to $$t$$, satisfying $$a_{n+1}-a_n\leq\frac12(a_n-a_{n-1})$$ and $$b_n-b_{n+1}\leq\frac12(b_{n-1}-b_n)$$. Then define $$T\colon\mathbb{Q}\to\mathbb{Q}$$ to be continuous, piecewise-linear map sending $$a_n\mapsto a_{n+1}$$ and $$b_n\mapsto b_{n+1}$$, and scales distance by $$\frac12$$ outside $$[a_1,b_1]$$, i.e., $$T(q)= \begin{cases} \frac12 (q-a_1)+a_2 & q\leq a_1\\ \frac12 (q-b_1)+b_2 & q\geq b_1\\ \dfrac{(a_n-q)a_n+(q-a_{n-1})a_{n+1}}{a_n-a_{n-1}} & q\in[a_{n-1},a_n]\\ \dfrac{(b_{n-1}-q)b_{n+1}+(q-b_n)b_n}{b_{n-1}-b_n} & q\in[b_n, b_{n-1}]. \end{cases}$$ By construction, $$T\colon\mathbb{Q}\to\mathbb{Q}$$ is bijective with $$\operatorname{Lip}(T)=\frac12<1$$ but has no fixed point (since $$t\notin\mathbb{Q}$$).