# 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.

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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.

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Itâ€™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.

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 (indeed, every uncountable Analytic set) ... To add on to the remark of the first paragraph, these facts can be proved by exhibiting the perfect set property. ZFC can prove that Analytic sets (which include the Borel sets) have the perfect set property. The axiom of choice shows there exists a set without the perfect set property. Large Cardinals or some projective determinacy shows that projective sets also have the perfect set property. – William Jun 16 '12 at 4:19 Nice proof that uses connectivity in every possible way. – Ewan Delanoy Jun 16 '12 at 4:35

(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.)

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 Aha, you were right to edit your original, mistaken claim. It is shown here that there is no continuous injective function ${\mathbb R}^2 \to {\mathbb R}$. – Ewan Delanoy Jun 16 '12 at 4:02