Shading in a simple closed curve I started thiking about this today and I have an answer I feel I can justify intuitively but not rigorously.
Let $\mathbb{S} = \{(x,y) \ | \ f(x,y)=c\}$ define the points on a simple closed curve (at least I think this is the right terminology; examples would be a circle, elipse or heart curve).
Is it always the case that set of points contained within the curve (i.e. the points that "shade it in") are equal to $\mathbb{S^*}=\{(x,y) \ | \ f(x,y) \lt c\}$
This seems to be true "intuitively" since you are making the $x, y$ values smaller and so $f(x,y)$ shrinks as well.  Can we perform some type of derivative on the curve to prove this?
Also, any help with terminology/notation would be greatly appreciated.  I think this falls under Topology but I'm not sure.
 A: Take $f(x,y) = \dfrac{1}{x^2 + y^2}$. The set $\{(x,y) : f(x,y) = 1\}$ is still the unit circle, but the region $\{(x,y) : f(x,y) < 1\}$ describes the outside of the circle.
A: This isn't the case if we just play around with the usual parametrisation of the circle as $\{x\in\mathbb{R}^2 \mid \|x\|=c\}$. If we instead define a curve to be $$\{x\in\mathbb{R}^2 \mid \|x\|^{-1}=c\}$$ then this defines a circle of radius $c^{-1}$ and the region $\{x\in\mathbb{R}^2 \mid \|x\|^{-1}<c\}$ is equal to $\{x\in\mathbb{R}^2 \mid \|x\|>c^{-1}\}$ which is the region we would normally call the exterior of the circle.
A: Note that taking $g(x,y) = 2 - f(x,y)$, we have $g$ equal to $1$ in exactly the same places as $f$, but the region where $f$ is less than one is precisely the region where $g$ is greater than one (and vice versa).
Also, discontinuous functions are perfectly legitimate functions! So actually if I tell you where $f$ takes the value $1$, I don't really tell you anything else about $f$. It's quite reasonable to have a function $f$ that is $1$ on the unit circle, then $7/8$ on a square around that, and everywhere else it's $\pi^{\pi^{7000}}$.
A: With a few additional assumptions your statement becomes true.
Suppose $f$ is continuous. You've assumed that $S$ is a simple closed curve, so by the Jordan curve theorem, the complement of $S$ in the plane consists of two components, one unbounded and one bounded. The bounded component is what you are calling the points inside the curve. Let's write $B$ for the bounded component and $U$ for the unbounded component.
Suppose that at least one point $(x,y)$ in $B$ satisfies $f(x,y)<c$. Then all points in $B$ satisfy this inequality, since $B$ is connected (definition of component), and $S$ by assumption contains all points with $f(x,y)=c$. And if one point in $U$ satisfies $f(x,y)>c$, then all points in $U$ do too (same reasoning).
So if (1) $f$ is continuous, and (2) one point in the bounded component satifies $f(x,y)<c$, and (3) one point in the unbounded component satifies $f(x,y)>c$, then your statement is true.
If any one of these assumptions is dropped, the statement no longer holds. Here's an example where (1) and (2) hold but not (3). Let $f(P)$ be the distance from $P$ to the origin if that distance is $\leq 1$. If the distance to the origin is $>1$, say $r$, then let $f(P)=1/r$. Then $S=\{P:f(P)=1\}$ is the unit circle, but the locus of $f(P)<1$ is everything else.
