What's the cardinality of all sequences with coefficients in an infinite set?

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 $\gamma\leq\alpha$ 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'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 $|A^\mathbb{N}| = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\cdot\aleph_0} = 2^{\aleph_0} = |A| < |\mathcal{P}(A)|$.

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 $\kappa^\lambda = \kappa$.

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.