Let $X$ be a metric space and let $f:X\to X$ be a contraction map. If $A$ is a proper subset of $X$, can $f(A)=X$? Intuitively, the answer should be no, but I can't see why. Is there something in Munkres or a basic text that answers questions like this about contraction maps?
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
The answer to your first question is yes, it can. Consider $X=\mathbb R$ and the function $f$ defined by $f(x)=\frac12x$ if $x\le1$ and $f(x)=\frac12|2-x|$ if $x\ge1$ (the graph of $f$ is a kind of SW-NE oriented zigzag with every slope equal to $\pm\frac12$). Then $f(A)=X$ with $A=X\setminus(0,2)$.
Another simple counterexample: $f(x) = \max(0, \frac x 2 - 1)$ is a contraction on $X = [0, \infty)$ which maps $[2,\infty) \subsetneq X$ to $X$.