Uncountable Pre-Image

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?

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No, some points may have countable, finite, even unique inverse images. – Gerald Edgar Apr 23 '13 at 0:30
Is there a relatively trivial example? I cannot seem to think of one... – John Bluto Apr 23 '13 at 0:51
Incidentally, there are continuous surjections $f: S^1\to S^2$ so that every preimage contains at most 3 points. (Actually, with countably many exceptions, preimage contains at most 2 points.) – Misha Apr 23 '13 at 2:17
You are right. I was thinking of a space-filling curve, which this is not. – Gerald Edgar Apr 23 '13 at 13:07

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.)
Actually, I only used the fact that the range of $f$ is open, not that $f$ is an open map. – Nik Weaver Apr 23 '13 at 1:00
Also, my argument actually shows that $f^{-1}(x)$ has cardinality $2^{\aleph_0}$, for those who care about such niceties. – Nik Weaver Apr 23 '13 at 1:01
Many thanks! One follow up question: How does your definition of U and V imply that the complement of $f^{-1}(x)$ is disconnected? – John Bluto Apr 23 '13 at 3:18
@John: The complement of $f^{-1}(x)$ is disconnected because it is the disjoint union of two nonempty open sets, $U$ and $V$. – Nik Weaver Apr 23 '13 at 5:00
Oh, one more comment. I wish I'd thought of saying this earlier. The proof that the complement of any countable subset of ${\bf R}^2$ is connected goes back to Cantor. I remember reading it in Dauben's biography. – Nik Weaver Apr 24 '13 at 3:20