Let:
$$ \begin{align} f(x_1, x_2, x_3, x_4) &= \sin(x_1 - x_2^3) + e^{x_4-x_3^4} + \tan(x_2)\ln(x_1)-100 \\ f &: X \to Y \\ X &\subset \mathbb{R}^4 \text{ is the natural domain of } f \\ Y &\subset \mathbb{R} \end{align} $$
Define:
$$ A = \{ \underline{x} \in X | f(\underline{x}) \lt 0 \} = f^{-1}(y<0) $$ $$ B = \{ \underline{x} \in X | f(\underline{x}) = 0 \} = f^{-1}(y=0) $$ $$ C = \{ \underline{x} \in X | f(\underline{x}) \le 0 \}= f^{-1}(y \le 0) $$
I want to check some topological properties of these sets.
I can see that $f$ is continuous wherever it's defined. $y<0$ is an open set, therefore $A$ is also an open set, as a preimage of an open set through a continuous function. Therefore, it can't be closed, since in an Euclidean space only the 2 trivial sets are clopen sets.
I've tried to prove something similar with B and C. The complement of $y=0$ is an open set, therefore, $B_{\neq} = f^{-1}(y\neq0)$ is an open set. But we can't conclude that B is closed, since the domain $X$ is an open set, and I assume $ B = {B_{\neq}}^c \cap X$.
- How should I think about complement sets where the domain is not the whole space?
- Should I be bothered by the fact that the domain isn't connected?
- I suspect B to be not open ("1 dimensional paths in $\mathbb{R}^4$") and not closed ("$x_1=0$ contains boundary points but isn't part of the domain or B"). I also suspect the same for C. but I don't know how to prove that without going into many many details.
- I think $A,B,C$, has the same interior, which is equal to $A$, and has the same boundary, which is not equal to $B$. Is that correct? Can we say anything else about the interior or boundary of these sets, $A, B, C$?
Partial answers and hints will also be appreciated.
My definitions:
- p is an interior point of set S if there exists an open ball centered at p which is completely contained in S
- S is an open set if all points in S are interior points of S
- The interior of a set S is the set of all the interior points of S
- p is an accumulation point of S if every neighborhood of p contains at least one point in S distinct from p
- S is a closed set if it contains all the accumulation points of S
- p is a boundary point of S if if every neighborhood of p contains both, at least one point in S distinct from p and at least one point not in S distinct from p
- The boundary of a set is the set of all the boundary points of S