Cardinality of the set of points on the $n$-unitsphere What's the cardinality of $S = \{ \mathbf x \in M^n : \|\mathbf x\|_2 = 1 \}$ for the sets $M = \mathbb Q, \mathbb R, \mathbb C$ – where $M^n$ is the n-ary Cartesian product $M \times \cdots \times M$ and $\|\mathbf x\|_2 := \sqrt{\displaystyle\sum_{k=1}^n |x_k|^2 }$?
 A: For $\mathbb{Q}$, it is $\aleph_0$. For $\mathbb{R}$ and above it is $\beth_1$. The same as $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ themselves.
All we have to do is show there is a subset of the sphere with such a cardinality. Then since the sphere is itself a subset of a set with that cardinality the whole sphere must have the same cardinality by "squeezing" ($a \le b$ and $b \le a$ so $a = b$, which also works for cardinals, at least in ZFC, which is essentially the "Cantor-Bernstein-Schroder theorem".)
Let the sphere be denoted $S^n_M$. Take a great circle in any plane, e.g. consider all points of the form $(x, y, 0, 0, 0, \cdots, 0) \in M^n$ with $\sqrt{x^2 + y^2} = 1$. Call this set $C^n_M$. Then this is just a circle in $M^2$ by the projection map $\mathrm{proj}: C^n_M \rightarrow M^2$ given by $\mathrm{proj}(a_1, a_2, \cdots, a_n) = (a_1, a_2)$.
For $M = \mathbb{Q}$, $\mathrm{proj}(C^n_M)$ is then easily seen to be the set of all rational points on the plane unit circle. This set is countably infinite, i.e. $C^n_\mathbb{Q} = \aleph_0$, because there are infinitely many rational points (they can be given by triples $(a, b, c)$ of nonnegative integers such that $\gcd(a, b, c) = 1$ and $c \ne 0$ and at least one of $a$ and $b$ also not 0. This is an infinite subset of countable $\mathbb{Z}^3$ so it is also countable.).
Now $M^n = \mathbb{Q}^n$ is countable as well (to construct the bijection is rather complicated). Thus, collecting that together, we have $|M| = \aleph_0$ and $|C^n_M| = \aleph_0$. Since $C^n_M \subset S^n_M \subset M^n$ we must have $\aleph_0 \le |S^n_M| \le \aleph_0$ thus $|S^n_M|$ is "trapped" between $\aleph_0$ and itself and so must be equal to it exactly.
For $M = \mathbb{R}$ we just get the unit circle for $C^n_M$. The unit circle has cardinality $\beth_1$ since one can give any point on it by its angle $\theta \in [0, 2\pi)$ from the x-axis and a real interval has the same cardinality as the reals themselves. Now $|M^n| = \beth_1$ as well by another complicated bijection (see here: https://math.stackexchange.com/a/183383/11172 for details of a special case, namely $n = 3$). Thus we should have again, since $C^n_M \subset S^n_M \subset M^n$ that $|C^n_M| \le |S^n_M| \le |M^n|$ which now gives $\beth_1 \le |S^n_M| \le \beth_1$ thus $S^n_M = S^n_\mathbb{R} = \beth_1$.
The same goes too for $M = \mathbb{C}$, although here to keep up the theme we need to be even more strict by saying that $x$ and $y$ must be specifically real numbers, and then again we get a unit circle and the same idea gives $\beth_1$ again.
A: Suppose that $n>1$. There is an important result to be aware of:
the cartesian product of an infinite set has the same cardinality.
From this, $\mathbb C$ and $\mathbb R$ are not so bad:
let $S^{n-1} \subseteq \mathbb R^n$ denote the unit sphere. Then if $S^{n-1}$ is infinite (which it is,) we $S^{n-1}-(0,\dots,1)$ (remove the north pole) has the same cardinality. In this case, consider the  map:
$$f:S^{n-1}-(0,\dots, 1) \to \mathbb R^{n-1}, \,\,\,\,\,\,\,\,\,(x_1, \dots,x_n) \mapsto \frac{1}{1-x_n}(x_1, \dots, x_{n-1}).$$
This is a bijection to $\mathbb R^{n-1}$, called the stereographic projection so the unit sphere has cardinality $|\mathbb R|$.
The same argument works for $\mathbb C^{n}$ by noting that $\mathbb C \sim \mathbb R^{2}$ as sets.
For $S^{n-1} \cap \mathbb Q^{n}$, we can say that it has cardinality at most $|\mathbb N|$, since there is an obvious injection into $\mathbb Q^n$ which is countable by the first reference. On the other hand, the solution set is not finite, since points of the form $(x/c, y/c, \cdots, 0)$ form an infinite family of solutions $x^2+y^2=c^2$.
