Completeness can be omitted from Banach Fixed Point Theorem? 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.
 A: 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.
A: "Completeness can't be omitted" is just saying that if $X$ is not complete, you can't be certain that a fixed point will exist for any given function. It doesn't mean you will never find one for specific functions.
It's like saying that "if $X$ is not complete, not all cauchy sequences converge". But you can definitely find some cauchy squences that do converge.
EDIT
It's now interesting to investigate how the theorem fails without completeness:
Firstly, here's an example of a contraction defined on a non-complete metric space with no fixed point:
$f:(0,1)\rightarrow(0,1)$ such that $f(x) = \dfrac{x}{2}$.
Clearly this is a contraction because $|f(x)-f(y)|=\Big|\dfrac{x}{2}-\dfrac{y}{2}\Big| = \dfrac{1}{2}|x-y|$ for any $x, y \in (0, 1)$. However, for any $x\in(0,1)$, $f(x)\neq x$, so there is no fixed point.
(On the natural domain of the function, the fixed point would be at $0$, but you can see that this is on the boundary of our non-complete set.)
Secondly, are there any cases where the metric space being incomplete results in more than one fixed point? The answer is no. Suppose $f:X\rightarrow X$ is a contraction and $X$ is any metric space (complete or not complete). Suppose $f$ has two fixed points at $x$ and $y$. Then $|f(x)-f(y)| = |x - y|$. This contradicts the definition of a contraction which requires $|f(a)-f(b)|<|a-b|$ for all $a,b\in X$.
