TITLE: Existence of an injective continuous function $\Bbb R^2\to\Bbb R$?
QUESTION [9 upvotes]: Let's say $f(x,y)$ is a continuous function. $x$ and $y$ can be any real numbers. Can this function have one unique value for any two different pairs of variables? In other words can $f(a,b) \neq f(c,d)$ for any $a$, $b$, $c$, and $d$ such that $a \neq c$ or $b \neq d$? I don't think there can at least not if the range of $f$ is within the real numbers. Could someone please offer a more formal proof of this or at least start me off in the right direction.

REPLY [2 votes]: Take any pair of distinct points $p,q$ of $\Bbb R^2$ and link them by two disjoint (except in the end points) arcs. For instance take $p=(1,0)$ and $q=(-1,0)$, linked  by the upper and lower halves of the unit circle. Now by hypothesis $f(p)\neq f(q)$, and the restriction of $f$ to either arc is a continuous function (of the parameter for the arc).  By the intermediate value theorem, every value in the interval between $f(p)$ and $f(q)$ is taken by $f$ on either of the arcs, but this contradicts the assumed injectivity.