# 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.

• – Martin Sleziak Apr 25 '17 at 11:59
• The negation of "for all ... then ... is true" is "there exists some ... such that ... is false", not "for all ... then ... is false". – Najib Idrissi Apr 25 '17 at 11:59
• Another way of phrasing Najib's comment is that theorems generally say "if these conditions are met, then this result holds". If the conditions aren't met, then the result may or may not hold; we can't tell. Kreyszig is asking you to find an example where the conditions aren't met and the result doesn't hold. You've found an example where the conditions aren't met and the result still holds. Which is cool, but not what is being asked. – Teepeemm Apr 25 '17 at 14:52
• Note that a contraction $T$ on any metric space $X$ extends in a unique way to a contraction on the completion $\bar X$, so there is always a unique fixed point $x_0 \in \bar X$. If $x_0 \in \bar X \setminus X$, then $T$ does not have a fixed point in $X$. – Justthisguy Apr 25 '17 at 16:28

"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 $a,b\in X$.

• O nice, would you like to give an example of such a function which fails to have unique fixed point in incomplete metric space. – Ajit Apr 25 '17 at 10:29
• Sure, you need something that tries to approach a fixed point on the boundary of the set. I'll edit my answer and add an example. – Harambe Apr 25 '17 at 10:38
• any function which has two or more fixed points due to incompleteness? – Ajit Apr 25 '17 at 10:59
• The answer is no - I'll add this in an edit too. – Harambe Apr 25 '17 at 11:12

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.