# Continuity of a function to the integers

I am trying to prove that in $\mathbb{Z}$ with co-finite topology the only path-connected components are the singletons. (I reckon that) showing that

"if a function $\gamma : [0,1] \to \mathbb{Z}$, where $\mathbb{Z}$ has co-finite topology, is continuous then it is constant"

should do the trick.

However I am not sure this is true, let alone if this is a good approach to the problem. Any thoughts about it?

Edit: I was thinking: suppose $x,y \in \mathbb{Z}$ and $\gamma : [0,1] \to \mathbb{Z}$, where $\mathbb{Z}$ has co-finite topology. Further suppose $x \neq y$, then $f^{1}(\mathbb{Z}\setminus \{x\})=(0,1]$ which is not open in $[0,1]$, contradicting continuity. Hence $x=y$. Does it seem right?

Edit 2: Forget the (stupid) edit above!

• This question was answered on MathOverflow: mathoverflow.net/questions/48970 Gowers's proof is quite nice. – Omar Antolín-Camarena Feb 7 '13 at 19:10
• Your proof is not correct: $(0, 1]$ is open in $[0, 1]$. Also if you assume $f(0) = x$ why can't we also have $f(.5) = x$? – Jim Feb 7 '13 at 19:16
• The edit is incorrect, it might be the case that many points are sent to $x$ and not just $0$. – Asaf Karagila Feb 7 '13 at 19:16

Suppose that $\gamma:[0,1]\to\Bbb Z$ is continuous but not constant. Then
$$\left\{\gamma^{-1}\big[\{n\}\big]:n\in\Bbb Z\right\}$$
is a non-trivial partition of $[0,1]$ into countably many closed sets. This MathOverflow question and its answers show that no such partition exists. For the sake of completeness I quote Tim Gowers’ very short answer: