Sets expressible as the image of exactly a countable number of functions. When working with a curve $C\subset\mathbb{R}^2$ one often finds that although $C$ can't be parametrized as a function $y(x)$ from the $X-$axis, it does admit a parametrization $x(y)$ from the $Y-$axis. A recurrent example is the cusp $x^3-y^2=0$;

It is easy to cook a curve which is not parametrized as a function from both the $X$ and $Y$ axis, but rather as a function from some rotated axis $X'$. Following this train of thought I started to think, for a given set $S\subset\mathbb{R}^2$, about the number of angles $\theta$ such that $S$ admits a parametrization from $X_\theta$, the rotated $X$ axis with angle $\theta$. This is equivalent to find the set
$$
\Theta_S:=\{\theta\in[0,\pi):P_\theta|_S\text{ is injective}\},
$$
where $P_\theta:\mathbb{R}^2\to X_\theta$ is the orthogonal projection onto $X_\theta$ (and '$|_S$' denotes restriction to $S$). Notice that the reason to restrict ourselves to angles in $[0,\pi)$ is to avoid repetitions (since if $S$ admits a parametrization from $X_\theta$ then it will also admit a parametrization from $X_{\theta+\pi}$).
For example, if(f) $S$ consists of only one point, then $\Theta_S=[0,\pi)$; if $S$ consists of two points then $\Theta_S=[0,\pi)\setminus\{\theta_0\}$ for some $\theta_0$;... In general it seems to be (at least) tedious to describe $\Theta_S$ even for finite $S$, so I focused in finding $|[0,\pi)\setminus\Theta_S|$ for finite $S$. Choosing points of $S$ in pairs and counting the number of total non-parallel lines we get the inequality
$$
|[0,\pi)\setminus\Theta_{S}|\leq\binom{|S|}{2},
$$
which (with a cardinality argument) in fact can be proven optimal. There are some interesting questions already here like "There exist, for every natural number $n$, a finite set $S_n$ such that $|[0,\pi)\setminus\Theta_{S_n}|=n?$" ---this problem feels very familiar to me so I would appreciate if someone could give me a reference (maybe from M. Aigner&G.Ziegler's Proofs from THE BOOK?)---, but I am relatively satisfied with the finite-case in comparison with the infinite-case:
When $S$ is infinite things get complicated, so let's think about familiar curves. If $S$ is a line then we have that $\Theta_S$ is $[0,\pi)$ missing one point, but this is a  highly degenerate case; if $S$ includes a $C^1$-arc with nonzero curvature then $\Theta_S\setminus[0,\pi)$ will have positive Lebesgue measure. In fact, the family of curves $\mathcal{S}=\{y=\alpha|x|\}_{\alpha>0}$ show that the measure of this set can be any number in $(0,\pi)$. It soon becomes clear that the interesting case here is to find sets $S$ such that $\Theta_S$ is small. For every closed path $\gamma$ in $\mathbb{R}^2$ we have $\Theta_\gamma=\phi$, and the cusp drawn above yields $|\Theta_C|=1$. Are there curves $C_n$ with $|\Theta_{C_n}|=n$? After playing for a while I've come up with a family of examples that accomplish this
$$
C_n=e^{iA_n},\hspace{.4cm}A_n:=\bigcup_{k=0}^{n-1}\left(\frac{2k\pi}{n},\frac{(2k+1)\pi}{n}\right).\text{ Graphically for $n=1,2,3,4$;}
$$


I was very satisfied when I found this so I started looking for a curve/set $N$ such that $|\Theta_N|=|\mathbb{N}|$. I first thought about some "limit-situation for $C_n$", and then something like $D=\{(x,\sqrt{1-x^2}):-1<x<1,x\in\mathbb{R}\setminus\mathbb{Q}\}$. I had no luck with the first idea, and $D$ can be proven to behave like $C_1$ by a cardinality argument.

Questions: Does such an $N$ even exist? If so, can someone provide an example, or at least an existence proof or some heuristic thoughts about it? Also I would love to see an example of a set $C$ with $|\Theta_C\setminus[0,\pi)|=|\mathbb{N}|$.

I started thinking of this just for fun and it soon became too intriguing to let it go. I am sorry for the extension but I didn't find anything online about this (I guess that nobody thought it could be of any use or, more likely, that it is just not a frequent search on the web; I would appreciate any reference in that case) so I definitely could not explain it in a shorter space. I hope it was readable. =)
 A: The answer to the finite case is yes, except for $n=2$. For $n=1$, simply pick two distinct points. For $n\geq 3$, one may take $S_n$ are as follows: choose any line $l$, pick $n-1$ distinct points along it and finally pick another point not on $l$.
EDIT: Actually, this answers the case for a set $C$ too. Pick a countably infinite set of distinct point on any one line $l$ and a point not on $l$.
A: An example of a set $N$ with $|\Theta_N|=|\mathbb{N}|$ is
$
N:=\{(x,0):x\not\in\mathbb{Z}\}\cup\{(0,1)\};
$
                  
In the picture the lines from $(0,1)$ through a hole $(k,0)$ of $N$ represent the directions of orthogonal projections $P_{\theta(k)}$ whose restriction to $N$ is injective. If $P$ is some orthogonal projection of $\mathbb{R}^2$ which is not of this form then either $P((0,1))=P((x,0))$ for some $x\in\mathbb{R}\setminus\mathbb{Z}$, or $P$ is the orthogonal projection with direction $(1,0)$ (and in this case it collapses all of $N$ but $(0,1)$). So the only orthogonal projections of $\mathbb{R}^2$ whose restriction to $N$ is injective are the $P_{\theta(k)}$'s, and hence
$$|\Theta_N|=|\{P_\theta(k):k\in\mathbb{Z}\}|=|\mathbb{N}|.$$
This example allows us to generalize with great flexibility. For instance, starting with some Cantor set $C\subset\mathbb{R}$, we can show the existence of $N_C\subset\mathbb{R}^2$ such that $\Theta_{N_C}$ has cardinal $c$ and yet its Lebesgue measure (in $[0,\pi)$) is $0$.
I've found this example because of @Fimpellizieri's answer. I just have been thinking about this enough time to notice that there exists kind of a duality between sets $S$ with "low and large" $|\Theta_S|$ (I seem to be unable to express this rigorously). Having this in mind it seems almost obvious the step from his answer to mine, so I am accepting his.
Added: I explained this question to a few friends of mine as a "geometric game" to see if they find some easy answer I missed, and one of them found this amazing one:
                             
Namely: Take the set $e^{i[0,\pi/2]}\cup\{(1/2,1/2)\}$ and remove a subset $P$ of $e^{i[\pi/6,2\pi/6]}$. Then the resulting set $M$ satisfies $|\Theta_M|=|P|$.
This solves the same problems that $N$ and is a bounded set; it was somehow annoying to me that I didn't have such an easy bounded set for the infinite case.
