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My motivation for asking this question is that a classmate of mine asked me some kind of question that made me think of this one. I can't recall his exact question because he is kind of messy (both when talking about math and when thinking about math).

I'm kind of stuck though. I feel like the set $A^{\mathbb{N}} = \{f: \mathbb{N} \rightarrow A, f \text{ is a function} \}$ should have the same cardinality as the power set of A, if A is infinite. On the other hand, in this post, it is stated that the sequences with real coefficients have the same cardinality as the reals.

It's easy to see that $A^{\mathbb{N}} \subseteq P(A)$, but (obviously) I got stuck on the other inclusion. Is there any general result that says anything else? References would be appreciated.

EDIT To clarify the intetion of this question: I want to know if there are any general results on the cardinality of $A^{\mathbb{N}}$ other that it is strictly less than that of the power set of A.

Also, I was aware that the other inclusion isn't true in general (as the post on here I linked to gave a counterexample), but thanks for pointing out why too. :)

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    Mathematics without choice is very very **very** weird. Just take a look at the wikipedia entry for AC, there's a short list of examples for things that happen without choice (I recently purchased Jech's book "The Axiom of Choice" in which there are many properties, theorems and proofs involving the negation of choice. Sounds mighty fun.)2010-12-15

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From Jech's Set Theory, we have the following theorems on cardinal exponentiation (a Corollary on page 49):

Theorem. For all $\alpha,\beta$, the value of $\aleph_{\alpha}^{\aleph_{\beta}}$ is always either:

  • $2^{\aleph_{\beta}}$; or
  • $\aleph_{\alpha}$; or
  • $\aleph_{\gamma}^{\mathrm{cf}\;\aleph_{\gamma}}$ for some $\gamma\leq\alpha$ where $\aleph_{\gamma}$ is such that $\mathrm{cf}\;\aleph_{\gamma}\leq\aleph_{\beta}\lt\aleph_{\gamma}$.

Here, $\mathrm{cf}\;\aleph_{\gamma}$ is the cofinality of $\aleph_{\gamma}$: the cofinality of a cardinal $\kappa$ (or of any limit ordinal) is the least limit ordinal $\delta$ such that there is an increasing $\delta$-sequence $\langle \alpha_{\zeta}\mid \zeta\lt\delta\rangle$ with $\lim\limits_{\zeta\to\delta} = \kappa$. The cofinality is always a cardinal, so it makes sense to understand the operations above as cardinal operations.

Corollary. If the Generalized Continuum Hypothesis holds, then $\aleph_{\alpha}^{\aleph_{\beta}} = \left\{\begin{array}{lcl} \aleph_{\alpha} &\quad & \mbox{if $\aleph_{\beta}\lt\mathrm{cf}\;\aleph_{\alpha}$;}\\ \aleph_{\alpha+1} &&\mbox{if $\mathrm{cf}\;\aleph_{\alpha}\leq\aleph_{\beta}\leq\aleph_{\alpha}$;}\\ \aleph_{\beta+1} &&\mbox{if $\aleph_{\alpha}\leq\aleph_{\beta}$.} \end{array}\right.$

So, under GCH, for all cardinals $\kappa$ with cofinality greater than $\aleph_0$ have $\kappa^{\aleph_0} = \kappa$, and for cardinals $\kappa$ with cofinality $\aleph_0$ (e.g., $\aleph_0$, $\aleph_{\omega}$), we have $\kappa^{\aleph_0} = 2^{\kappa}$. (In particular, it is not the case the cardinality of $A^{\mathbb{N}}$ is necessarily less than the cardinality of $\mathcal{P}(A)$).

Then again, GCH is usually considered "boring" by set theorists, from what I understand.

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    If we were to replace $2^{\aleph_\beta}$ (in the first dot point) with $\gimel(\aleph_\beta)$, would the theorem still hold? Where $\gimel \kappa$ is just shorthand for $\kappa^{\mathrm{cf}(\kappa)}$; I'm using notation I saw [here](http://en.wikipedia.org/wiki/Gimel_function).2014-05-02
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Take $A = \mathbb{R}$ then you have:

$|A| = 2^{\aleph_0}$ and so $|A^\mathbb{N}| = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\cdot\aleph_0} = 2^{\aleph_0} = |A| < |\mathcal{P}(A)|$.

Addendum:

To the comment, as well as a small "obvious" remark that should probably be mentioned - if $\kappa>1$ and $\lambda\ge\aleph_0$ are two cardinal numbers, to "generate" a relatively small (although this is a bad term, as it is consistent that this size is pretty much unbounded) cardinal which is invariant under powers of $\lambda$, one can always take $\kappa^\lambda$. The above computation is as good for this case to show that the result has this property.

It is also obvious that if $\kappa$ is already invariant under powers of $\lambda$ then $\kappa^\lambda = \kappa$.

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    Andy, very well. I have added an obvious remark, that probably has very little use for a development of intuition, but it pretty much states all there is to say about cardinals which are invariants under exponentiation of $\aleph_0$ (or any other cardinal for that matter).2010-12-15
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Arturo Magidin's answer has the general theorem. Here are two more facts that can be useful:

  • If $\aleph_0 \leq \lambda$ and $2 \leq \kappa \leq \lambda$ then $\kappa^\lambda = 2^\lambda = |P(\lambda)|$

  • If $\aleph_0 \leq \lambda \leq \kappa$ then $\kappa^\lambda = |\{ X \subseteq \kappa : |X| = \lambda \}|$

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If $A$ is countably infinite, then the cardinality of $A^{\Bbb N}$ equals the cardinality of the reals: $|A^{\Bbb N}|=|A|^{|\Bbb N|}=\aleph_0^{\aleph_0}\leq (2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\aleph_0}=2^{\aleph_0}=|\Bbb R|$ and $\aleph_0^{\aleph_0}\geq 2^{\aleph_0}$ together gives $|A^{\Bbb N}|=|\Bbb R|$.

See the Wikipedia page on cardinal numbers.

As shown by Asaf in his answer, $|A^{\Bbb N}|=2^{\aleph_0}$ if $|A|=2^{\aleph_0}$. It follows easily that if $\aleph_0\leq|A|\leq2^{\aleph_0}$, then still $|A^{\Bbb N}|=2^{\aleph_0}$.

When $A>2^{\aleph_0}$, things start to become complicated, and depends on whether you assume the Generalised Continuum Hypothesis or not. See for example these notes by Charles Morgan.