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Consider the implicit function defined by the equation $$ f(x,y) = 0 \tag{1} \label{eq1} $$ where $f : \mathbb R^2 \rightarrow \mathbb R$ is some continuous function. Suppose that the implicit function defined by \eqref{eq1} partitions $\mathbb R^2$ into two disjoint sets, $\mathbb R^2_+$ and $\mathbb R^2_-$, such that $$ \mathbb R^2 = \mathbb R^2_+ \cup \mathbb R^2_- $$ I've found that, for some examples of $f(x,y)$, I am able to show that for any pair $(x_0,y_0) \in \mathbb R^2$, if $$ f(x_0,y_0) > 0 $$ then $(x_0,y_0)$ belongs to either $\mathbb R^2_+$ or $\mathbb R^2_-$ (but not both), and if $$ f(x_0,y_0) < 0 $$ then $(x_0,y_0)$ belongs to the other disjoint set. The simplest example of this involves the equation of a circle with radius $1$, which is $$ x^2 + y^2 - 1 = 0 $$ If, for any $(x_0,y_0) \in \mathbb R^2$, $$ x_0^2 + y_0^2 - 1 > 0 $$ then $(x_0,y_0)$ is outside the circle, and if $$ x_0^2 + y_0^2 - 1 < 0 $$ then $(x_0,y_0)$ is inside the circle. However, I am unable to prove that this pattern holds more generally for all continuous functions $f$. How can one go about showing this? Does there exist a theorem that addresses this?

mhdadk
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    Maybe there is some connectedness assumption on your partitioned sets. Otherwise $x^2 - y^2 = 1$ can still be treated as a set partitioning $\Bbb{R}^2$. Jordan curve theorem or Jordan-Schoenflies theorem could be related anyway. – onRiv Feb 03 '22 at 12:42
  • I think that, in general, given a function $f:\Omega\subset \mathbb R^2\to\mathbb R$, you can look at the monotonicity of the map $y\mapsto f(x,y)$ for $x$ fixed and the limits of the restricted function $f(x,y)\vert_{x=x_0}$ at the boundary of its domain of definition, in order to find two regions $\Omega_1\cup \Omega_2$ in which $f$ implicitly defines two function $\phi_j:\Omega_j\subset \mathbb R\to\mathbb R$, with $j=1,2$. – Vajra Feb 03 '22 at 14:24

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