Can a spiral, $r=f(n)\theta$, be defined so that all points in a plane are mapped bijectively to an angle?
Related:
Consider $r=(1/n)\theta$. Is it possible to define the ratio of points in a plane to points along this spiral as $n\to \infty$.
Motivation/Background:
I'm a high school math teacher - confident with multivariable calculus but no experience with real analysis. Question was a passing thought on a slow day.
The number of points in the plane and the number of points in the spiral path both have the same uncountably infinite cardinality. It is possible to put them in one-to-one correspondence, but the particular attempt at a one-to-one correspondence you have in mind is another matter altogether.
Consider the points that lie along a particular ray from the origin. For example, consider the positive $x$-axis. For any particular value of $n$, if $f(n) > 0$ the intersection of this set of points with the spiral occurs at angles of the form $\theta=2k\pi$ or $\theta = -\pi-2k\pi$ for any non-negative integer $k$, that is, $\theta = \ldots, -5\pi,-3\pi, -\pi, 0, 2\pi, 4\pi,\ldots.$ (The negative values come from the fact that the polar coordinates $(-f(n)\pi, -\pi)$ and $(f(n)\pi, 0)$ identify the same point.) For any real value of $n$, there are only countably many of those points, but there are uncountably many points on the $x$-axis. Since this is true for any real $n$, it's true in the limit as $n \to \infty$.
On the other hand, if $f(n)$ is a continuous function, non-zero for some positive value of $n$, such that $\lim_{n\to\infty} f(n) = 0$ (for example, $f(n) = 1/n$), any particular point in the plane (other than the origin) will be on the spiral $r = f(n)\theta$ for (countably) infinitely many values of $n$. But for every such value of $n$, there is a larger value of $n$ for which that same point is not on the spiral $r = f(n)\theta$. So I would not say that that the spiral $r = f(n)\theta$ maps an angle to this point "in the limit" as $n\to \infty$, or even that this point is on the spiral in the limit. The same reasoning applies to every point in the plane, so no point other than the origin is on the spiral in the limit.
As I already mentioned, it is possible to explicitly construct a bijective mapping from the real line $\mathbb R$ to the plane $\mathbb R^2$. This answer describes a bijective mapping between the interval $(0,1]$ and the region $(0,1]\times(0,1]$, and the "appendix" of that answer indicates how to make a bijection between $(0,1]$ and $\mathbb R$, which also gives you the tools to make a bijection between $(0,1]\times(0,1]$ and $\mathbb R^2$. But the resulting bijection is not a continuous function, and moreover there cannot be a continuous bijection between the two sets.
Space-filling curves are continuous mappings from a line segment to a region of the plane, but they are not bijective.
Note that for any $n$, the area of the spiral is zero, whereas the area of the unit disk around the origin is $\pi$.
