Existence of an injective continuous function $\Bbb R^2\to\Bbb R$? 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.

No, because the continuous image of a connected set is connected. If $f: \mathbb{R}^{2} \to \mathbb{R}$ is a continuous bijection, then we can consider $f( \mathbb{R}^{2} \setminus \{ 0, 0 \} ) = \mathbb{R} \setminus \{ f(0, 0) \}$. Let $c = f(0, 0)$. Since $\mathbb{R}^{2} \setminus \{ (0, 0) \}$ is connected, so must be $\mathbb{R} \setminus \{ f(0, 0) \} = ( - \infty, c) \cup (c, + \infty)$, which it obviously isn't. Thus either $f$ is not continuous, or it is not a bijection.
EDIT: Commenter pointed out that the OP only supposed $f$ was injective. A similar method holds. Again, $f (\mathbb{R}^{2})$ is a connected subset of $\mathbb{R}$, and so is an interval, since all connected subsets of $\mathbb{R}$ are intervals (though its endpoints may be infinite). So suppose that the interval $(a, b) \subseteq f(\mathbb{R}^{2})$. Let $a < c < b$, and let $\mathbf{x} = (x_{1}, x_{2})$ be such that $f(\mathbf{x}) = c$. Then again, $\mathbb{R}^{2} \setminus \{ \mathbf{x} \}$ is connected, so $f( \mathbb{R}^{2} \setminus \{ \mathbf{x} \} )$ is connected. But we can consider that $$f( \mathbb{R}^{2} \setminus \{ \mathbf{x} \} ) = ( (- \infty, c ) \cap f(\mathbb{R}^{2} \setminus \{ \mathbf{x} \})) \cup ( (c, + \infty) \cap f(\mathbb{R}^{2} \setminus \{ \mathbf{x} \})) ,$$
and so is not connected. Thus either $f$ is not injective, or its not continuous.
