Does there exist an injective function $f:\mathbb R^2 \to \mathbb R$ such that $f$ is continuous in one of the variables $x$ or $y$? Does there exist an injective function $f:\mathbb R^2 \to \mathbb R$ such that $f$ is continuous in one of the variables $x$ or $y$ ? I only know that such an injection cannot be continuous in each real variable $x$ and $y$ . Please help . Thanks in advance 
 A: There is no such function. Let us assume that for each $x$, the function $f_x (y) = f(x,y)$ is continuous the other case is analogous. Note that each $f_x$ is injective.
An injective, continuous function $f: \Bbb{R} \to \Bbb{R}$is necessarily strictly increasing (or decreasing) and it's image is an open interval, so that $I_x = f_x (\Bbb{R})$ is an open interval.
By injectivity, the $(I_x)_{x \in \Bbb{R}}$ are pairwise disjoint.  But since $\Bbb{R}$ is second countable, there can be no uncountable pairwise disjoint collection of nonempty open sets in $\Bbb{R}$.
Alternative argument: Each $I_x$ contains some rational $q_x$ and the $q_x$ are pairwise distinct. This is impossible since the rationals are countable.
A: No. Here is an outline:
Suppose $g:\mathbb{R} \to \mathbb{R}$ is continuous and injective, then $g(\mathbb{R})$ is a non trivial interval.
Suppose $I_\alpha$ is an uncountable collection of non trivial intervals. Then at least two of the intervals overlap.
Now consider the uncountable non trivial collection $\{f(\mathbb{R},y) \}_y$
