Could the set of fixed points be uncountable?

Question

Let $f $ be a continuous function defined from $[0,1] $ to $[0,1] $ and

$A=\{x\in [0,1] \; : \; f (x)=x\} $.

We assume that

$$\forall (a,b)\in [0,1]^2 \;\;$$ $$a\ne b\implies \;\;\exists x\in [a,b] \;:\; x\notin A .$$

Could $A $ be uncountable?

With $f (x)=x|\sin (\frac {1}{x}) |$ for $x\ne 0$ and $f (0)=0$,

we have an example where $A $ is infinite and countable, but I have no example with $A$ uncountable.

Answer 1

The complement of $A$ is open and dense in $[0,1]$. Ignoring the endpoints, we can write it as countable union of countably many pairwise disjoint open intervals $(a_i,b_i)$, $i\in\Bbb N$. Given any such enumeration of pairwise disjoint open intervals, we can define $$f(x)=\begin{cases}(x-a_i)(b_i-x)+x&\text{if }x\in(a_i,b_i)\text{ for some }i\in\Bbb N\\x&\text{otherwise}\end{cases} $$ Then $f$ is continuous and has precisely the complement of the open intervals we started with (i.e., $A$) as its fix point set.

Now all you need to find is a suitable $A$, i.e., an uncountable closed subset of $[0,1]$ with dense complement. The Cantor set is a suitable example.

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We could also make $f$ monotonous, or $C^\infty$ (or both). Just by replacing the interval-wise quadratics with something more suitable.

Answer 2

If $C\subseteq[0,1]$ is any closed set such that $0,1\in C$, then there is a continuous map $f:[0,1]\to[0,1]$ whose fixed point set is $C$. Indeed, the complement of $C$ is an open subset of $(0,1)$ and thus is a disjoint union of open intervals $(a_n,b_n)$. If we choose continuous functions $f_n:[a_n,b_n]\to[a_n,b_n]$ for each $n$ such that $f_n(a_n)=a_n$, $f_n(b_n)=b_n$ and $f_n(x)\neq x$ for all $x\in (a_n,b_n)$, we can then define $f:[0,1]\to[0,1]$ by $f(x)=x$ if $x\in C$ and $f(x)=f_n(x)$ if $x\in(a_n,b_n)$, and $f$ will be continuous (continuity of $f$ at points of $C$ takes a bit of work to check; I leave that as an exercise) and the fixed point set of $f$ is $C$.

For the existence of such functions $f_n$, note that if $a_n=0$ and $b_n=1$, then we could take $f_n(x)=x^2$. For general $a_n$ and $b_n$, you can then conjugate this by a homeomorphism between $[0,1]$ and $[a_n,b_n]$.

So, in particular, for instance, if $C\subseteq[0,1]$ is the Cantor set, there is a continuous map $f:[0,1]\to[0,1]$ with $C$ as its fixed point set. Since the Cantor set is uncountable and contains no interval, this satisfies your requirements.

Answer 3

Let $K$ denote the Cantor set. Recall that $K$ is compact, uncountable, and contains no interval of positive length. Define $f(x) = x +(1-x)d(x,K),$ where $d(x,K)$ is the distance from $x$ to $K.$ Then $f$ maps $[0,1]$ continuously to $[0,1],$ and $f(x)=x$ iff $x\in K.$