2

As part of a larger inquiry, I suspect and am trying to prove the following : Let $\phi$ be a continuous function defined on $[0,1]$. Then $\phi^{-1}(0)$ has an at most countable number of connected components.

My guess is that the argument is about its complement being open as $\phi$ is continuous. Therefore, unless there's at most one connected component, there is a nonzero distance between each of them. Somehow, I suspect I could prove from that that we can count the connected components (e.g. by increasing order) - which seems obvious to the intuition.

[EDIT] : Motivation for this : I'm trying to prove that a piecewise-$\mathcal{C}^1$ path in the complex plane can always be described in "polar" coordinates ($\phi(t)=r(t)e^{i\theta(t)}$) where $\theta$ is continuous. Therefore, I wanted to simply define this function $\theta$ by induction by giving it a different expression everytime $\text{Arg} (\phi(t))$ crosses the line at which $\text{Arg}$ is discontinuous.

James Well
  • 1,209
  • It might help to see, just a little bit, what this larger enquiry is. Maybe there's some more context, some stronger hypotheses on $\phi$, to help prove/disprove this conjecture – FShrike Jun 12 '22 at 22:01
  • One could look at the standard space-filling curve from $[0,1]$ onto $[0,1]\times[0,1]$ and then compose with the projection onto either axis; it's either a good candidate for a counterexample, or a good example to look at to see why the countability is forced. – Greg Martin Jun 12 '22 at 22:15

1 Answers1

5

Let $C$ be the Cantor set. This is totally disconnected so it has uncountbaly many components. But $C$ is $\phi^{-1}(0)$ where $\phi (x)=d(x,C)$.

[Every singleton $\{x\}$ in $C$ is a connected component: Let $A$ be the connected component containing $x$. If there is a point $y$ other than $x$, say $y<x$, in $A$ then there is a point $z\in (y,x)$ which is not in $C$ (because $C$ contains no open interval). Now $A=[A\cap (-\infty,z)]\cup [A\cap (z\infty)$ makes $A$ disconnected. This contradiction shows that $A=\{x\}$. We have proved that each singleton $\{x\}$ in $C$ is a connected component, so there are uncountably many connected components].

geetha290krm
  • 36,632
  • @James Well I have added a detailed proof of the fact that there are uncountably many connected components]. – geetha290krm Jun 13 '22 at 05:05
  • So, in particular, every closed subset of $[0,1]$ can be realized as the preimage of a continuous map? – Kenta S Jun 13 '22 at 06:30
  • 2
    In any metric space any closed set $C$ is the pre-image of $0$ under the map $x \to d(x,C)$ and this map is always continuous. @KentaS – geetha290krm Jun 13 '22 at 06:32
  • 1
    Cool! So with that insight, the question is essentially asking whether there exists a closed subset $C\subset[0,1]$ with uncountably many connected components. – Kenta S Jun 13 '22 at 06:57