# Topology: Boundary of a set

Consider a set $$D = \{ x \in \Bbb R^n\mid h(x) < 0 \}$$ where $$h:\Bbb R^n \to\Bbb R$$ is a continuous function.

On what condition, can I say that the set $$\{ x \in \Bbb R^n\mid h(x) = 0 \}$$ is the boundary of $$D$$? Or, equivalently, $$\{ x \in \Bbb R^n | h(x) \leqslant 0 \}$$ is the closure of $$D$$?

My initial guess is that the solution of $$h(x) = 0$$ only has to form a curve. But is it enough? I need a rigorous mathematical condition.

Thank you!

• Would the condition "interior of $\{ x \mid h(x) = 0 \}$ is empty" be acceptable? Commented May 9, 2020 at 22:51

(1) $$D=h^{-1}(-\infty,0)$$, which is inverse of open set, is open.
When $$x_n\in D\rightarrow x$$, then $$h(x)=\lim\ h(x_n)$$ since $$h$$ is continuous. Hence $$h(x)\leq 0$$ That is, $$\overline{D} \subseteq \{ x| h(x)\leq 0\}$$
(2) Note that the inclusion is proper : Consider a function $$h$$ whose restriction to $$\{x||x|\leq 1\}$$ is $$h(x)=|x|$$.
(3) Since $$D$$ is open so any point in $$D$$ is an interior point in $$D$$. Further, $$x\in \overline{D} - D$$ can be an interior point in $$\overline{D}$$ : $$h(x)=-|x|$$