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What is the result of a natural number power the cardinality of an infinite set? Is it the cardinality of the infinite set?

Thank you!

Hayden
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user12342494
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  • That's simply $\infty$. As far as I understand, you're mixing between Cardinal Arithmetic and Natural Arithmetic. Here is a good answer which explain this. – barak manos Jun 13 '16 at 14:46
  • If you are referring to things like $2^{\aleph_0}$, depending on the status of the continuum hypothesis, it would be $\aleph_1$. I.e. $2^{|\Bbb N|}=|\Bbb R|$ (note that $|\Bbb R|\neq |\Bbb N|$. They are both infinite, but different sizes of infinite). See this page. – JMoravitz Jun 13 '16 at 14:50
  • Yes it is about that. What is for example $3^\aleph_0$? – user12342494 Jun 13 '16 at 14:55
  • @user12342494: $3^{\aleph_0}$ is the number of different functions from $\mathbb N$ to a fixed $3$-element set. This happens to be equal to $2^{\aleph_0}$ or indeed to $n^{\aleph_0}$ for any finite $n$. – hmakholm left over Monica Jun 13 '16 at 15:00
  • OK. Thank you very much! – user12342494 Jun 13 '16 at 15:05
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    @JMoravitz's comment is potentially misleading. $2^{\aleph_0}$ always equals $|\mathbb R|$; whether this cardinality is also $\aleph_1$ is what the Continuum Hypothesis is about. – hmakholm left over Monica Jun 13 '16 at 15:07
  • @JMoravitz: Your comment is also misleading in the sense that $e^5$ is a perfectly valid answer to the question "what is the number $x$ for which $\ln x=5$?"; in the same manner, $\aleph_2^{2^{\aleph_{42}}}$ is a perfectly valid answer to a question about cardinal arithmetic, even if we can't quite say for which $\alpha$ it is equal to $\aleph_\alpha$. Not to mention that we have the $\beth$ numbers to help and compensate for this a little bit (although not enough, admittedly). – Asaf Karagila Jun 13 '16 at 20:26

2 Answers2

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(Assuming by "natural number power the cardinalty of an infinite set you mean $n^{|X|}$)

Recall that given two cardinals $\kappa,\lambda$, the cardinal $\lambda^\kappa$ is the cardinality of the set of all functions $\kappa \to \lambda$.

Let $X$ be an infinite set with cardinality $\kappa$. The value of $n^\kappa$ for a natural number $n$ depends on $n$, coming down to three cases:

  • $n=0$, in which case $0^\kappa=0$ as there is no function from a non-empty set to $0=\emptyset$.
  • $n=1$, in which case $1^\kappa=1$ as there is exactly one function from a non-empty set to $1=\{\emptyset\}$.
  • $n>1$, in which case $n^\kappa=2^\kappa$ is strictly larger than $\kappa$. The equality $n^\kappa=2^\kappa$ for $n\geq 2$ is because $2\leq n \leq 2^\kappa$ and monotonicity $$2^\kappa \leq n^\kappa \leq (2^\kappa)^\kappa =2^{\kappa^2}=2^\kappa$$ along with the result that $\kappa^2=\kappa$ for $\kappa$ infinite by AC.

If, on the other hand, you mean $\kappa^n$, then using the Axiom of Choice it can be shown that $\kappa^2=\kappa$, as above, and so by induction $\kappa^n=\kappa$ for all $n\geq 1$. Naturally, $\kappa^0=1$.

Hayden
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The question you ask is answered by the following theorem:

Theorem. For cardinal numbers $\mathfrak{a}$ and $\mathfrak{b}$ such that $1 < \mathfrak{a} \leq b$ and $\mathfrak{b}$ is transfinite, we have:

$\mathfrak{a}^{\mathfrak{b}} = 2^{\mathfrak{b}}$

The proof of the theorem is a fairly easy consequence of the following two lemmas.

Lemma 1. If $\mathfrak{a}$ is a cardinal such that $\mathfrak{a}^2 = \mathfrak{a}$, then $\mathfrak{a} = 2 \mathfrak{a} = 3 \mathfrak{a} = \mathfrak{a}^2$.

Lemma 2. If $\mathfrak{a}$ is a transfinite cardinal, then $\mathfrak{a}^2 = \mathfrak{a}$.

The proof of Lemma 2, in turn, depends essentially on the Axiom of Choice.

murray
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