Vrftjkry

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. :)

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 $gammaleqalpha$ where $aleph_{gamma}$ is such that $mathrm{cf};aleph_{gamma}leqaleph_{beta}ltaleph_{gamma}$.

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 $langle alpha_{zeta}mid zetaltdeltarangle$ with $limlimits_{zetatodelta} = kappa$. 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
$$aleph_{alpha}^{aleph_{beta}} = left{begin{array}{lcl}
aleph_{alpha} &quad & mbox{if $aleph_{beta}ltmathrm{cf};aleph_{alpha}$;}\
aleph_{alpha+1} &&mbox{if $mathrm{cf};aleph_{alpha}leqaleph_{beta}leqaleph_{alpha}$;}\
aleph_{beta+1} &&mbox{if $aleph_{alpha}leqaleph_{beta}$.}
end{array}right.$$

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'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 $kappa^lambda = |{ X subseteq kappa : |X| = lambda }|$
  

Take $A = mathbb{R}$ then you have:

$|A| = 2^{aleph_0}$ and so $|A^mathbb{N}| = (2^{aleph_0})^{aleph_0} = 2^{aleph_0cdotaleph_0} = 2^{aleph_0} = |A| < |mathcal{P}(A)|$.

Addendum:

To the comment, as well as a small "obvious" remark that should probably be mentioned - if $kappa>1$ and $lambdagealeph_0$ 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 $kappa^lambda = kappa$.

If $A$ is countably infinite, then the cardinality of $A^{Bbb N}$ equals the cardinality of the reals: $|A^{Bbb N}|=|A|^{|Bbb N|}=aleph_0^{aleph_0}leq (2^{aleph_0})^{aleph_0}=2^{aleph_0aleph_0}=2^{aleph_0}=|Bbb R|$ 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 $aleph_0leq|A|leq2^{aleph_0}$, 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.