My question today is about the Adeles and whether they have the same cardinality as the real numbers.

Certainly $\mathbb{R}$ has more elements than $\mathbb{Z}$. This is by Cantor diagonalization. Here are some interesting twists I have read on that discussion (not important now)

We also know that p-adic numbers $\mathbb{Q}_p$ has the same cardinality as $\mathbb{R}$. You can imagine that allowing infinitely many digits to the left of the decimal point should be the same as allowing them to the right.

What happens when we multiply all of them? I made the mistake of writing an equality here:

$$ \mathbb{A} \neq \prod \mathbb{Q}_p$$

Instead the Adeles are a "restricted product" of all p-adic numbers over all primes $p = 2,3,5,7,11,13\dots$ :

$$ \mathbb{A} = \Big\{ (x_2, x_3, x_5, \dots,x_p, \dots ) \in \prod \mathbb{Q}_p \Big| \text{ most of } x_p \in \mathbb{Z}_p \Big\} $$

Is the "size" or "number of" Adeles stricly larger than $\mathbb{R}$? Certainly the straight product would be bigger:

$$ |\mathbb{A}| \leq \left|\prod \mathbb{Q}_p \right| = \left| \mathbb{R}^\mathbb{Z} \right| $$

Since we only have the restricted product... how big is this set exactly? How does it compare to more familiar sets like $\mathbb{R}, \mathbb{Z}$ or $2^\mathbb{Z}$ ?


Cardinal arithmetic tells us $|\mathbb{R}^\mathbb{Z}|=|\mathbb{R}|^{|\mathbb{Z}|}=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\times \aleph_0}=2^{\aleph_0}=|\mathbb{R}|$.

And $|\mathbb{R}|\le|\mathbb{A}|\le|\mathbb{R}^\mathbb{Z}|=|\mathbb{R}|$ implies $|\mathbb{A}|=|\mathbb{R}|=\mathfrak{c}$.

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