How $f_1(x,y,z)=c_1$ and $f_2(x,y,z)=c_2$ result in a curve in $\mathbb R^3$?

Question

Though a usual way to represent a curve or an arc in space is by particularization, there is also another way to define a curve in space.

Introduction:

A general curve in $\mathbb R^2$ can be represented as $f(x,y)=c$. If we go to the $\mathbb R^3$ and want to have the same curve, we can say the equation for the mentioned curve is $f(x,y,z)=c$ in which $z=0$ or the system of equations: $c_1=f_1(x,y,z):=f(x,y)=c$ and $c_2=f_2(x,y,z):=z=0$. In this case (a curve on $x-y$ plane) turned to be the solution for the two-equations $f_1(x,y,z)=c_1 and f_2(x,y,z)=c_2$. Another example is considering the $x-$axis as a curve: ${\{(x,y,z)\in \mathbb R^3 | \ y=z=0}\}$ is also in the form of the two-equations $f_1(x,y,z)=c_1$ and $f_2(x,y,z)=c_2$. I am trying to prove then intuitively understand why any general curve is equivalent to the two-equations $f_1(x,y,z)=c_1$ and $f_2(x,y,z)=c_2$, but I can't though I tried.

However, I know (intuitively not a proof) for the case of a surface which I write it here which helps to clarify my question: I claim that the single equation $f(x,y,z)=c$ is a surface in $\mathbb R^3$. Equation $f(x,y,z)=c$ means that for a given $x$ and $y$ we can find the $z$ or $z$'s; this in $\mathbb R^3$ means for any point on $x-y$ plane (:= a given $x$ and $y$), we spot one or many points projected to $z-$axis. In the case of a curve on $x-y$ plane, for a given $x$ we find one or many $y$ which is same as interpretation of $f(x,y)=c$.

My questions are:

1- How interpretation of solving the two-equations $f_1(x,y,z)=c_1$ and $f_2(x,y,z)=c_2$ is a description of a curve in $\mathbb R^3$?

2- If solving the two-equations $f_1(x,y,z)=c_1$ and $f_2(x,y,z)=c_2$ results in a curve in $\mathbb R^3$, how to prove this mathematically?

3- In connection with question 2 (but a simpler case) consider $f(x,y)=c$. We say it is curve when we can prove these:

A) For a given $x$, there is one or more $y$ to fit in $f(x,y)=c$. How to prove that $y$'s are 'separate points' (i.e. they are not making an interval, so $f(x,y)=c$ is a curve and not a surface or volume or so on)?

B) How to prove for any pair $(x_1, y_1)$ fit into $f(x,y)=c$ in real numbers and for some arbitrary $\epsilon>0$, there are infinitely many pairs of $(x_i, y_i)$, such that the 'arc length' between $(x_1, y_1)$ and any of $(x_i, y_i)$'s are less than $\epsilon$?

Thank you.

Answer 1

Another intuitive way to look at the same question is by counting degrees of freedom. In $\mathbb{R}^3$ you need 3 coordinates to uniquely specify a point. If you have two constraining equations $f_1(x,y,z)=0$ and $f_2(x,y,z)=0$ then you're down 2 degrees of freedom, meaning that you need 1 coordinate to uniquely specify a point. That's your curve right there.

Answer 2

You might be looking for the preimage theorem, which says that if $f\colon X\to Y$ is a smooth map between smooth manifolds and $y\in Y$ is a regular value of $f$, then $f^{-1}(y)$ is a smooth submanifold of $X$ of codimension $\dim Y$.

In your case $X=\mathbb R^3$, $Y=\mathbb R^2$ and $f = (f_1,f_2)$. So whenever $(c_1, c_2)$ is a regular value of $f$, i.e. the jacobian of $f$ is of full rank at all preimages of $(c_1,c_2)$, you get a smooth $1$-dimensional manifold $$f^{-1}(c_1,c_2) = \left\{\, x\in\mathbb R^3 : f_1(x)=c_1, f_2(x)=c_2 \,\right\}.$$