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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. :)

Community
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Andy
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  • In general $\mathcal{P}A \not\subseteq A^\mathbb{N}$... intuitively, because you can't label all the elements of $A$ by only countably many labels! – Katriel Dec 15 '10 at 17:31
  • Maybe you rather feel like the set ${\mathbb{N}}^A$ should have the same cardinality as the power set of A, if A is infinite? – Hendrik Vogt Dec 15 '10 at 18:16
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    The containment $A^{\mathbb N}\subseteq{\mathcal P}(A)$ is false. There is a difference between the collection of countable subsets of $A$ and the collection of (countable) sequences from $A$. The difference is irrelevant under the axiom of choice for $A$ infinite, but may matter in contexts where choice is not assumed. – Andrés E. Caicedo Dec 15 '10 at 18:43
  • @Andres Caicedo: thanks for pointing that out. I'm used to assume AC, so I rarely imagine how to do things without it (also because I'm not that good still). What was in my mind was something like this: $A^{\mathbb{N}} \subseteq A^A \cong P(A)$ by "$\cong$" I mean "in bijection with". – Andy Dec 15 '10 at 19:03
  • 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.) – Asaf Karagila Dec 15 '10 at 19:14

4 Answers4

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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.

Arturo Magidin
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    Arturo, as usual a great answer, I'd mention that the cofinality is always a cardinal number as well. As per the last remark about the interest of GCH, in the introduction part of one of the books I read ("Introduction to Cardinal Arithmetic" by three German guys, which I don't recall) they say that if you accept GCH then you are pretty much done with cardinal arithmetic and there is nothing in the book for you ;) – Asaf Karagila Dec 15 '10 at 18:24
  • @Asaf: Thanks; I'll add the note that the cofinality is always a cardinal. – Arturo Magidin Dec 15 '10 at 18:27
  • 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. – goblin GONE May 02 '14 at 03:15
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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 $\kappa^\lambda = \kappa$.

Asaf Karagila
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  • Yes, exactly. That's the point of my question: since there is an easy counterexample to my intuition, I was wondering if there were any results that said something specific about the cardinality of this set. Since just saying that $\left|A^{\mathbb{N}}\right| < \left|P(A)\right|$ isn't much of a result (see my remark about the inclusion). Sorry if I wasn't really clear. – Andy Dec 15 '10 at 17:31
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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). – Asaf Karagila Dec 15 '10 at 18:33
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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 $\kappa^\lambda = |\{ X \subseteq \kappa : |X| = \lambda \}|$

Carl Mummert
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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.

Konrad Swanepoel
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