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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?
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 $t\notin\mathbb{Q}$).
