Does there exist a function whose graph contains any finite collection of points in the plane, no two with the same x coordinates, up to translation? Given any finite set of points in $\mathbb{R}^{2}$, no two with the same $x$-coordinates, it is easy to find a function $f:\mathbb{R}\rightarrow\mathbb{R}$ whose graph passes through all of them. In fact, Lagrange interpolation tells us functions as nice as polynomials will do the job. Can we do it with just one function?
The obvious answer is no, since if $f(x)=y$ then any set of points containing $(x,y)$, $z\neq y$, will never be covered by $f$. But what if we are allowed to translate this function not only horizontally but also vertically? Does there exist a function $f$ such that no matter what finite set of points we start with, some set of points in $f$ contains the &quot;shape&quot; of those finite points?
My first thought was that of the pathological solutions to the Cauchy functional equation $f(x+y)=f(x)+f(y)$. But I don't know how to prove this does/doesn't work, and as a sort of bonus I'd prefer something which is more axiomatically efficient (this certainly doesn't look like a problem requiring even all of $\sf{ZF}$).
I have a suspicion the answer may be no, however. My reasoning, intuitively, is that there are uncountably infinitely many sets of finite points, even up to translation. Each collection of (at least $2$) finite points has nonzero width. From here I wish to use an argument similar to that which states every collection of pair-wise disjoint intervals in $\mathbb{R}$ is countable, where we note that even if we were to try and go &quot;finite set by finite set&quot; and fit the function to one finite set, then to the next, and so on, we would only cover countably many finite sets (since $\mathbb{Q}\subset\mathbb{R}$ is dense and countable). Of course the issue here is that the function may satisfy many finite sets in the same interval, and in fact we can have uncountably infinitely many finite sets satisfied in any given interval.
I would not be surprised if this were possible. I would be somewhat surprised, however, if it were still true should we add certain smoothness conditions on the function. Perhaps differentiability, or infinite differentiability. Or maybe just continuity. To prevent the question from becoming too broad, I'll simply ask: Is it possible with no restrictions on $f$? Is it still possible if $f$ is continuous?
For real $(a,b,c)$ the simple example of a parabola with symmetry axis parallel to $x-$ axis:
$$y= a+\sqrt{ b x +c}$$
For real $(a,b,c)$ the simple example of a parabola with symmetry axis parallel to $x-$ axis:
$$y= a+\sqrt{ b x +c}$$