Always a value with uncountably many preimages? (for a continuous real map on the plane) Let $f$ be a continuous map ${\mathbb R}^2 \to {\mathbb R}$. For $y\in {\mathbb R}$, denote by $P_y$ the preimage set $\lbrace (x_1,x_2) \in {\mathbb R}^2 | f(x_1,x_2)=y \rbrace$.
Is it true that
(1) At least one $P_y$ is uncountable ?
(2) At least one $P_y$ has the same cardinality as $\mathbb R$.
Some easy remarks :


*

*(2) is stronger than (1).

*(2) follows from (1) if we assume the GCH.

*If there is a point $(x_0,y_0)$ such that the partial derivatives $\frac{\partial f}{\partial x}(x_0,y_0)$ and $\frac{\partial f}{\partial y}(x_0,y_0)$ exist and one of them is nonzero, then 
(2) (and hence (1)) follows from the implicit function theorem.
 A: Consider the restriction of $f$ to $\mathbb{R}_x=\mathbb{R} \times \{x\}$ for fixed $x$. If its image is a point, we're done. Otherwise, its image is an interval in $\mathbb{R}$, and that interval contains a subinterval with rational endpoints. Since there are only countably many such intervals, there must be one that's contained in the images of uncountably many $\mathbb{R}_x$; then any point in that interval has uncountable preimage.
A: It’s well-known that every uncountable closed subset of $\Bbb R^n$ (indeed, every uncountable Borel set) has cardinality $2^\omega$, so (1) and (2) are equivalent, since every $P_y$ is closed. In fact a strong form of (2) is true.
(2) is certainly true if $f$ is constant. If not, $\operatorname{ran}f$ contains an open interval $(a,b)$. For each $y\in(a,b)$, $\Bbb R^2\setminus P_y$ must be disconnected. But for any countable set $S\subseteq\Bbb R^2$, $\Bbb R^2\setminus S$ is arcwise connected and therefore connected, so $|P_y|=2^\omega$. 
To see that $\Bbb R^2\setminus S$ is arcwise connected, fix $p,q\in\Bbb R^2\setminus S$ with $p\ne q$. There are $2^\omega$ straight lines through $p$, so most of them miss $S$, and similarly for $q$. Thus, there are straight lines through $p$ and $q$ that miss $S$ and intersect.
A: (1) Deleted my answer for this part. 
(2) If you have that there exists a $P_y$ which is uncountable, then $P_y$ has the cardinality of $\mathbb{R}$ even without the continuum hypothesis. This is because since $f$ is continuous, $P_y$ is a closed set since $y$ is closed. By the Cantor Bedixson theorem, it can be written as a union of a countable set and a perfect set. Perfect sets have cardinality of the continuum. (I guess they are called perfect because they are not counterexamples to the continuum hypothesis.)
