TITLE: Topology: Boundary of a set
QUESTION [5 upvotes]: 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!

REPLY [2 votes]: (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
$\overline{D}$ : $h(x)=-|x|$