$f: \mathbb{R}^2 \to \mathbb{R}$ a continuous open map, show that for each $x \in \text{range}(f)$, $f^{-1}(x)$ is always uncountable.

Question

Let $f: \mathbb{R}^2 \to \mathbb{R}$ be a continuous open map. Show that in fact for each $x \in \text{range}(f)$, $f^{-1}(x)$ is always uncountable.

I know that if this was simply a projection onto the one of the two coordinates, this problem would be trivial, since it being continuous, surjective, and open would imply that it's a quotient map, and taking the preimage would yield an uncountable sequence of ordered pairs. However, it instead being a continuous open map, with no other information, seems to make it slightly more complicated.

Couldn't we use the fact that the irrationals (which are uncountable) are dense in the reals implies, for $ A = \mathbb{R} \setminus \mathbb{Q}$ and every open subset of $\mathbb{R}$, say $B$, that $A \cap B \neq \emptyset$, so if we partition the outcome space into disjoint subsets whose union is $\mathbb{R}$, making $f$ into a surjective and therefore quotient map onto it's range, then the inverse function is surjective and continuous, so we can determine that the yielded sequence of ordered pairs is uncountable due to the density of the irrationals in the range and, by continuity and surjectiveness, the domain?

Would this line of thinking be correct? If someone could correct me if I am wrong or formalize their own solution that would be great! Thank you in advance.

Answer 1

$\Bbb R^2$ is connected, so $f[\Bbb R^2]$ must be a connected, open subset of $\Bbb R$: an open interval, an open ray, or $\Bbb R$ itself. These are all homeomorphic to $\Bbb R$, so without loss of generality you can assume that $f$ is surjective. (If not, just compose it with a homeomorphism from its range onto $\Bbb R$.) Now fix $a\in\Bbb R$, and let $F=f^{-1}\big[\{a\}\big]$, $U=f^{-1}\big[(\leftarrow,a)\big]$, and $V=f^{-1}\big[(a,\to)\big]$. $U$ and $V$ are open, $F$ is closed, and $\{U,V,F\}$ partitions $\Bbb R^2$. Fix $u\in U$ and $v\in V$. If $F$ is countable, there is a path from $u$ to $v$ in $\Bbb R^2\setminus F$. Indeed, the path can be a union of two line segments: just pick lines $\ell_u$ through $u$ and $\ell_v$ through $v$ that avoid $F$ and have different slopes, let $p$ be their point of intersection, and use the segments $\overline{up}$ and $\overline{pv}$. Avoiding $F$ is possible because only countably many slopes have to be avoided. (This argument actually requires only that $|F|<|\Bbb R|$.)

Let $P=\overline{up}\cup\overline{pv}$. Then $P$ is compact and connected, so $f[P]$ is a closed interval in $\Bbb R$. But $f(u)<a<f(v)$, so $a\in f[P]$, and $P\cap F\ne\varnothing$, a contradiction.

Answer 2

I don't think that your idea with rationals vs. irrationals can lead anywhere.

Consider $y\in f(\mathbb R^2)$ and $x\in f^{-1}(y)$. As $\mathbb R^2$ is open, its image is an open neighbourhood of $y$. Hence there exist $x_1,x_2\in \mathbb R^2$ with $f(x_1)<y<f(x_2)$. Let $u\in\mathbb R^2$ be orthogonal to $x_2-x_1$. For $a\in\mathbb R$ consider the continuous function $g_a\colon[0,1]\to\mathbb R^2$ $$ t\mapsto (1-t)x_1+tx_2+at(1-t)u.$$ For each $a$, the intermediate value theorem fives us a $t\in(0,1)$ with $f(g_a(t))=y$. Note that $g_a(t)=g_{a'}(t')$ with $a\ne a'$ is only possible if $t=t'\in\{0,1\}$, hence we ibtain a different point for each of the uncountably many $a\in\mathbb R$.