TITLE: Completeness can be omitted from Banach Fixed Point Theorem?
QUESTION [9 upvotes]: In Kreyszig's Functional Analysis, page no. 303, exercise no. 3 says that completeness cannot be omitted from Banach's Fixed Point Theorem. But if we take $f(x)=x^2$ from an incomplete metric space $(-1/3,1/3)$ to $(-1/3,1/3)$, here $f$ is a contraction and this $f$ has a unique fixed point $0$. Here I have omitted completeness but I am still getting a unique fixed point - where am I wrong?
Banach's Fixed Point Theorem:
Consider a non-empty metric space $X = (X, d)$. Suppose that $X$ is
complete and let $T: X \to X$ be a contraction on $X$. Then $T$ has precisely
one fixed point.

REPLY [6 votes]: As well as Shanye2020's counterexample on an open interval, we can construct a counterexample on a closed subset of the rationals: take $M=[-\frac13,\frac13]\cap\mathbb Q$, and $f(x)=x^2+\frac16$. Then $f(x)$ has a fixed point in $[-\frac13,\frac13]$ as a subset of $\mathbb R$, but this fixed point is irrational.