I've been reading about space filling curves, and been asking myself this question. 


If $f: \mathbb{R}^{2} \rightarrow \mathbb{R}$ is a continuous open map, is it true that $\forall x \in$ range$(f)$ , $f^{-1}(x)$ is always uncountable? 



Uncountable Pre-Image
I think so, yes. Let $x$ be in the range of $f$ and define $U = f^{-1}((-\infty, x))$, $V = f^{-1}((x,\infty))$. Since $f$ is open, $U$ and $V$ are both nonempty. So they are disjoint nonempty open sets, which means that the complement of $f^{-1}(x)$ is disconnected. But the complement of any countable subset of ${\bf R}^2$ is connected. (Even path connected, by a simple cardinality argument: for any distinct $p$ and $q$ one can find uncountably many paths from $p$ to $q$, any two of which are disjoint except for the points $p$ and $q$. So a countable set can obstruct only countably many of them.)


