Is it possible for the graph of a multivariable function to be a curve instead of a surface? The graph of a function, let's say for example $f: \mathbb{R}^2  \rightarrow \mathbb{R} $ is a subset of the Cartesian product of the domain and the codomain, in this case of $\mathbb{R}^3$. When speaking about the graph of functions like $f$ the term that is always associated is that one of surface. However i was wondering if there is a possibility that the graph of a function of this type is instead a curve, that can be represented as a vector function $\vec{f}$. For example, a function $F(x, y)$ with the property that the condition of each level set is only satisfied by one point.
$$ F(x, y) = c$$
I imagine that the previous example is not possible, nonetheless i haven't been able to prove it. I don't know if the question is rather trivial, and i have a misconception about the concept of the  graph of a function.
 A: It's a well known result that $\mathbb{R}^2$ has the same cardinality as $\mathbb{R}$, so there is a bijection $h:\mathbb{R}^2\longrightarrow\mathbb{R}$. Then, for each $c\in\mathbb{R}$ there exists one, and only one point $(x,y)\in\mathbb{R}^2$ such that $h(x,y)=c$.
Anyways, the graph of such a function is obviously very difficult to imagine. So probably if you ask almost any condition over $F$ such as continuity, monotony, or whatever, that won't be possible.
A: Unless I'm overlooking something, the graph in ${\mathbb R}^3$ of $z = f(x,y),$ where
$$ f(x,y) \; = \; \sqrt{x^2 + y^2 - 1} \; + \; \sqrt{1 - x^2 - y^2} $$
is the circle in the $xy$-plane with radius $1$ and center $(0,0,0).$
Note that the domain of $f(x,y)$ is $x^2 + y^2 = 1$ (a circle in the $xy$-plane, not a cylinder in $xyz$-space), and for each point $(a,b)$ such that $a^2 + b^2 = 1$ we have $f(a,b) = 0.$
A: I answer here your question in my comment. If we add continuity we cannot construct such a function (the one with level sets consisting of just one point). Let $F:\mathbb{R}^2\longrightarrow\mathbb{R}$ be a continuous function, and let $(x_1,y_1),(x_2,y_2)\in\mathbb{R}^2$. Applying Intermediate Value Theorem to the single variable functions:
$$g_1(t):=\begin{cases}F((x_1,y_1)+2t\cdot(x_2-x_1,0)) &\text{ if }t\in[0,\frac{1}{2}] \\F((x_1,y_1)+(2t-1)\cdot(0,y_2-y_1)) &\text{ if } t\in(\frac{1}{2},1]\end{cases}\\
g_2(t):=\begin{cases}F((x_1,y_1)+2t\cdot(0,y_2-y_1)) &\text{ if }t\in[0,\frac{1}{2}] \\F((x_1,y_2)+(2t-1)\cdot(x_2-x_1,0)) &\text{ if } t\in(\frac{1}{2},1]\end{cases}$$
We get that for all $y_1\in(g_1(0),g_1(1))$ there exists $c_1\in(0,1)$ such that $g_1(c_1)=y_1$, and for all $y_2\in (g_2(0),g_2(1))$ there exists $c_2\in(0,1)$ such that $g_2(c_2)=y_2$. Note that $g_1(0)=g_2(0)$ and $g_1(1)=g_2(1)$ because of how we defined them, then for every: $$y\in(g_1(0),g_1(1))=(g_2(0),g_2(1))=(F(x_1,y_1),F(x_2,y_2))$$ there exists $c_1,c_2\in(0,1)$ such that $g_1(c_1)=g_2(c_2)=y$. The existence of this implies that there are: $$(a,b)\in L((x_1,x_2),(x_2,y_1))\cup L((x_2,y_1),(x_2,y_2)) \\ (c,d)\in L((x_1,y_1),(x_1,y_2))\cup L((x_1,y_2),(x_1,y_2))$$
(where $L((u,v),(w,z)):=\{(u,v)+t\cdot (w-y,z-v) : t\in[0,1]\}$ is the segment joining two points).
Such that $F(a,b)=y=F(c,d)$. This makes impossible what we want.
