Boundary points of a domain bounded by a continuous curve Suppose $F:\mathbb{R^2}\to \mathbb{R}$, which is given by $F(y_1,y_2)=\frac{y_1^2}{4}+\frac{y_2^2}{9}-1$. $S=\{(y_1,y_2) |F(y_1,y_2)=0\}$, and $D=\{(y_1,y_2) |F(y_1,y_2)<0\}$. 
I want to show $\partial D=S$, where $\partial D$ denotes all boundary points of $D$.
By our definition of boundary point, I need to show if $x=(x_1,x_2)\in \mathbb{R^2}$ is a boundary point, then $\forall r>0$, $B_r(x) \cap D \neq \emptyset$ and $B_r(x) \cap D^C \neq \emptyset$. I can show by continuity of $F$, $D$ and $D_1=\{(y_1,y_2) |F(y_1,y_2)>0\}$ are open, hence points inside them can't be boundary points. Then let $x \in S$, how can I show inside every open ball $B_r(x)$, there are points $z$, s.t. $F(z)$ is bounded away from $0$?
This fact seems trivial if I view it geometrically on a plane, but I can't transform it into $\epsilon$-$\delta$ argument.
 A: You already have shown that $\partial D\subset S$, using the continuity of $F$.
For the converse inclusion, consider a point $x=(x_1,x_2)\in S$, and let an $r>0$ be given. You don't have to show that $F$ is "bounded away from $0$" in $B_r(x)$, but that $F$ assumes both negative and nonnegative values in $B_r(x)$.
Since $x\in B_r(x)$, and $F(x)=0$, it remains to produce a point $y\in B_r(x)$ with $F(y)<0$. To this end put
$$\delta:=\min\left\{1,{r\over 2|x|}\right\}$$ and let $$y:=(1-\delta) x\ .$$
Then $$|y-x|=\delta|x|\leq{r\over2}<r\ .$$ On the other hand, using $F(x)=0$ we obtain
$$F(y)=(1-\delta)^2\left({x_1^2\over4}+{x_2^2\over9}\right)-1=(1-\delta)^2\bigl(F(x)+1\bigr)-1=-\delta(2-\delta)<0\ .$$
Answering a comment of the OP:
What is $\partial S$? It is a simple fact that $\partial(\partial A)=\partial A$ for any set $A$, and this implies $$\partial S=\partial(\partial D)=\partial D=S\ .$$
But it is not difficult to prove $\partial S=S$ directly: Given an $x\in S=\partial D$, any neighborhood of $x$ intersects $S$ and $D\subset\complement S$. It follows that $S\subset\partial S$. Conversely: Assume $x\notin S$. Then either $F(x)<0$ or $F(x)>0$. In both cases the continuity of $F$ implies the existence of a neighborhood $B_r(x)$ that does not intersect $S$. It follows that $x\notin\partial S$.
A: Not true in general, here a counter example:
$\Bbb D=\{(x,y) \ / \ x^2+y^2\leq 1\}$ The unit disk.
$F(x,y)=d((x,y),\Bbb D)$ is continuous and 
$S=\Bbb D$  ,  $D=\emptyset$ and $\partial D\neq S$.
A: The gradient of $F(y_1,y_2)$ is $\nabla F=(y_1/2,2y_2/9)$. This is the zero vector only at the origin, which does not lie on the ellipse. Therefore at any point $P$ on the ellipse you can find a unit vector $\vec{u}(P)$  such that the directional derivative
 $\nabla_\vec{u}F(P)\neq0$. Thus moving a tiny bit back/forth in the direction of $\vec{u}$ shows that the function $F$ has both negative and positive values in an arbitrarily small neighborhood of $P$.
Using the directionally derivative essentially reduces this to the univariate case, where it is easy to show that if $g(c)=0$ and $g'(c)\neq0$, then a function $g(x)$ has both negative and positive values in any neighborhood of $c$.
