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The question is pretty simple:

Can the cardinality of a set $\mathbb{S}^n$ be written in terms of $\lvert \mathbb{S}\rvert$? This includes transfinite cardinals.

Here, $\mathbb{S}^n=\{(x_1,\cdots,x_n):x_i\in \mathbb{S}\}$ and $n\in\Bbb N$.

I’ve researched the answer and I can’t find anything. I understand set theory just fine so don’t hold back on any explanations or proofs. A well-justified answer would get the $\color{green}\checkmark$.

gen-ℤ ready to perish
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    If $\mathbb{S}$ is finite then $|\mathbb{S}^n| = |\mathbb{S}|^n$, otherwise $|\mathbb{S}^n| = |\mathbb{S}|$. In the latter case, we must appeal to Axiom of Choice to obtain the result. – Akira Jun 18 '19 at 00:54
  • This link may be of your interest https://math.stackexchange.com/questions/2916466/if-x-is-infinite-then-x-and-x-times-x-are-equinumerous – Akira Jun 18 '19 at 01:00
  • @LeAnhDung That’s a $\color{green}\checkmark$-worthy response (but also, explanation would be appreciated). – gen-ℤ ready to perish Jun 18 '19 at 01:23
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    @LeAnhDung If $n$ is finite then $|S^n|=|S|^n$ regardless of whether $S$ is finite or infinite, and the axiom of choice is not involved; it's the definition of $|S|^n$. Assuming the axiom of choice, if $n$ is finite and $|S|$ is infinite, then $|S|^n=|S|$. – bof Jun 18 '19 at 01:28
  • A tangential comment on the previous comments, concerning the axiom of choice. The statement that, for all infinite $S$, $|S^2|=|S|$ is actually equivalent to the axiom of choice. – Andreas Blass Jun 18 '19 at 02:55

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I assume that by $n$, you mean a natural number $n\in\mathbb N$.

If $|\mathbb S|=m$ is finite, this is simply a matter of combinatorics: for each coordinate in the $n$-tuple $(x_1,\dots,x_n)$ we have $m$ options, so the total number of $n$-tuples is simply $n\cdot m$.

If $\mathbb S$ is infinite, things get a little more complicated. For $\aleph$ numbers we have that $\aleph_\alpha\cdot\aleph_\alpha=\aleph_\alpha$. You can prove this by using the canonical well-order on $Ord^2$ by letting $(\alpha,\beta)<(\gamma,\delta)$ iff

  • $\max\{\alpha,\beta\}<\max\{\gamma,\delta\}$, or
  • $\max\{\alpha,\beta\}=\max\{\gamma,\delta\}$ and $(\alpha,\beta)$ is less than $(\gamma,\delta)$ using the lexicographical order.

If there exists a bijection between $\mathbb S$ and an aleph number $\aleph_\alpha$, then we can see that $\mathbb S^n$ has cardinality $|\mathbb (\aleph_\alpha)^n|=\prod_n \aleph_\alpha=\aleph_\alpha$. When the Axiom of Choice is true, any infinite $\mathbb S$ has a cardinality equal to an aleph number.

However, without choice, there exist sets $\mathbb S$ that have a cardinality unequal to any $\aleph$ number, and hence we cannot say $\mathbb S^n$ has the same cardinality as $\mathbb S$. In fact without the Axiom of Choice it is consistent that $|\mathbb S|\neq |\mathbb S\times\mathbb S|$. The statement that $\mathbb S$ and $\mathbb S\times \mathbb S$ have the same cardinality is sufficient to prove the Axiom of Choice.

With or without choice, you could define the cardinality of $\mathbb S^n$ as the cardinality of the set of functions $f:\{0,1,\dots,n-1\}\to\mathbb S$. This is how $|\mathbb S^n|$ is defined. Since $n=\{0,1,\dots,n-1\}$ is a cardinal itself, and cardinal exponentiation $\kappa^\lambda$ is defined as the cardinality of the set of functions $\lambda\to\kappa$, we see that $|\mathbb S^n|=|\mathbb S|^n$.

Vsotvep
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