Circles and surfaces We say a subset $\overline{D}$ of $\Bbb R^2$ is $x\textbf{-normal}$ is there exist $a,b\in \Bbb R$ with $a<b$ and continuous functions $\psi_1,\psi_2:[a,b]\to \Bbb R$ such that $\forall x\in ]a,b[(\psi_1(x)<\psi_2(x))$ (open interval as a domain) and $\overline{D}=\{(x,y)\in \Bbb R^2: a\leq x\leq b \ \wedge \psi_1(x)\leq y\leq \psi_2(x)\}$.
We say $\overline{D}$ is $y\textbf{-normal}$ if the set $D:=\{(x,y)\in \Bbb R^2:(y,x)\in \overline{D}\}$ is $x$-normal.
We say $\overline{D}$ is normal if it is both $x$-normal and $y$-normal.
We say $\overline{D}$ is a normal regular domain if it is normal and we require the inequality $\psi_1<\psi_2$ to also hold on $\{a,b\}$.
Question : I want to prove that any circle is a normal regular domain. I already proved that any circle is normal. However the regularity escapes me.
I suppose it's easy to generalize from the circles centered at the origin to any circle, so I'll write what I've done on that.
Suppose the circle $C:=\{(x,y)\in \Bbb R^2:x^2+y^2\leq r^2\}$ is a normal regular domain.
Then twe have $C=\{(x,y)\in \Bbb R^2: a\leq x\leq b \ \wedge \psi_1(x)\leq y\leq \psi_2(x)\}$ for some $a,b\in \Bbb R$ with $a<b$ and some continuous functions such that $\forall x\in [a,b](\psi_1(x)<\psi_2(x))$.
Since $(a,\psi_1(a)),(a,\psi_2(a)),(b,\psi_1(b)),(b,\psi_2(b))\in C$ we have $a^2+(\psi_1(a))^2\leq r^2$ and $a^2+(\psi_2(a))^2\leq r^2$. Subtracting the inequalities from each other we can see that $|\psi_1(a)|=|\psi_2(a)|$ and because of the inequality $\psi_1<\psi_2$ we must have $\psi_1( a) =-\psi_2(a)$.
I can't do anything from this nor do I see any other way.
 A: The definition of normal regular domain you give is not what I would expect. And the circle does not satisfy it. In your definition the boundary of a regular domain should contain some segment.
A: Following up on your proof by contradiction, prove that $[a,b\textbf{]}=[-r,r\textbf{]}$.
$\subseteq$: Let $x\in [a,b\textbf{]}$. Since $C$ is regular $(x,\psi_1(x))\in C$. Therefore $x^2\leq x^2+(\psi_1(x))^2\leq r^2$ and $-r\leq x\leq r$ follows.
$\supseteq:$ Let $x\in [-r,r\textbf{]}$. Now note that $r^2-x^2\ge 0$ and
$$\begin{align}
  -r\leq x\leq r &\Longrightarrow &x^2\leq r^2 \\
  &\Longrightarrow &x^2+r^2-x^2\leq r^2 \\
  &\Longrightarrow &x^2+(\sqrt{r^2-x^2})^2\leq r^2\\
  &\Longrightarrow &\Bigl(x,\sqrt{r^2-x^2}\Bigr)\in C
\end{align}$$
Therefore $x\in [a,b\textbf{]}$.
From $[a,b\textbf{]}=[-r,r\textbf{]}$ conclude that $a=-r$ and use the fact that $(a,\psi_1(a)),(a,\psi_2(a))\in C$:
$$\begin{align}
  (a,\psi_1(a)),(a,\psi_2(a))\in C&\Longrightarrow a^2+(\psi_1(a))^2\leq r^2 \wedge a^2+(\psi_2(a))^2\leq r^2\\
&\Longrightarrow r^2+(\psi_1(a))^2\leq r^2 \wedge r^2+(\psi_2(a))^2\leq r^2\\
&\Longrightarrow \psi_1(a)=0=\psi_2(a)
\end{align}$$
which contradicts the assumption that $\psi_1(a)<\psi_2(a)$.
$\therefore$ C isn't regular.
