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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!

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  • $\begingroup$ Would the condition "interior of $\{ x \mid h(x) = 0 \}$ is empty" be acceptable? $\endgroup$
    – md2perpe
    Commented May 9, 2020 at 22:51

1 Answer 1

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(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 $\overline{D}$ : $h(x)=-|x|$

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