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 $\|\mathbf x\|_2 := \sqrt{\displaystyle\sum_{k=1}^n |x_k|^2 }$?

  • $\begingroup$ The answer depends on $n$ as well. For $\mathbb R$, the unit sphere is only $2$ points, while for $\mathbb C$, it is uncountable. For higher dimensional things, it is uncountable, though. $\endgroup$ – Andres Mejia Apr 9 '17 at 6:53

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.

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  • $\begingroup$ I would say you made an argument for your claims, but you didn't really show anything. For example, the empty set with cardinality 0 is trivially a subset of every set. But all you can say is that the cardinality of the "bigger" set is at least the cardinality of the empty set. If you could edit your answer and provide explicit bijections from S to M – or at least cardinal arithmetic – it would be very much appreciated. $\endgroup$ – Markus Klyver Apr 8 '17 at 12:30
  • $\begingroup$ @Markus Klyver: The trick is also that it is at most the same cardinality. In other words, it is both at least and at most. To do an explicit bijection from a one-dimensional set to a polydimensional one is rather complicated and would bloat the post, so it is easier to just leverage cardinal comparisons. For $M = \mathbb{Q}$, $C$, the circle section has cardinality $\aleph_0$, $M^n = \mathbb{Q}^n$ has cardinality $\aleph_0$, and $C \subset S \subset M^n$ thus $\aleph_0 \le |S| \le \aleph_0$ so $|S| = \aleph_0$. $\endgroup$ – The_Sympathizer Apr 9 '17 at 2:02
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    $\begingroup$ But one could/should point out that the squeezing is a proper theorem which has a name (Cantor--Schröder--Bernstein), so people know where to look. $\endgroup$ – martin.koeberl Apr 9 '17 at 2:48

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 $x^2+y^2=c^2$.

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