Cardinality statements involving the power set of an infinite set.

Question

Working on the question

$\quad$ How many subsets of $\mathbb{N}$ have the same cardinality as $\mathbb{N}$?

leads to another question (and expanding the ZF axiomatic system is welcomed here)...

Note: I did a precursory search to see if this question has been asked before on this site and nothing stood out. Also, see the comments found after this question.

============================================

For what infinite sets $\Bbb S$ does the following hold true?

Let

$\quad \mathcal I(\mathbb S) = \{A \in \mathcal P(\mathbb S) \, \mid \, A \text{ is infinite} \} $

Then $|\mathcal I(\mathbb S)| = \mathcal |\mathcal P(\mathbb S)|$.

Moreover, if

$\quad \mathcal G(\Bbb S) = \{A \in \mathcal P(\mathbb S) \, \mid \, |A| \lt |\mathbb S| \}$

then $|\mathcal G (\Bbb S)| \lt \mathcal |\mathcal P(\mathbb S)|$;
or more strongly stated, $|\mathcal G (\Bbb S)| = |\mathbb S|$.

Answer 1

In ZF, without the axiom of choice, there is not a lot we can say for arbitrary $\mathbb S$. In fact, it is possible for stuff like: $|\mathcal I(\mathbb S)|<|\mathcal G(\mathbb S)|=|\mathcal P(\mathbb S)|-1<|\mathcal P(\mathbb S)|=|\mathcal I(\mathbb S)|+|\mathcal I(\mathbb S)|$ to happen (This is the case for amorphous sets).

With the axiom of choice, only the first question can be answered, and the answer is "yes":

$|\mathcal I(\mathbb S)| = \mathcal |\mathcal P(\mathbb S)|$

Clearly $|\mathcal I(\mathbb S)|≤|\mathcal P(\mathbb S)|$, so we only need to find an injection from the powerset to the "infinite"set. I will prove something stronger, I will show that there is injective from $\mathcal P(\mathbb S)$ to $[\mathbb S]^{|\mathbb S|}=\{x\in\mathcal P(\mathbb S)\mid |x|=|\mathbb S|\}$.

In ZFC we have that for every infinite cardinality $κ$, we have $2\kappa=κ$, in particular, $|\mathbb S|=2|\mathbb S|$, take $A\subseteq \mathbb S$ such that $|A|=|\mathbb S\setminus A|=|\mathbb S|$, and so $|\mathcal P(A)|=|\mathcal P(\mathbb S)|$, and for each $x\in \mathcal P(A)$, then $|x\cup (\mathbb S\setminus A)|=|\mathbb S|$, you can show that indeed $f(x)=x\cup (\mathbb S\setminus A)$ is injective from $\mathcal P(A)$ to $[\mathbb S]^{|\mathbb S|}$.

$|\mathcal G (\Bbb S)| \lt \mathcal |\mathcal P(\mathbb S)|$

If $|\mathbb S|=κ^+$ for some infinite $κ$, then ZFC proves that $|\mathcal G (\Bbb S)|=2^κ$. Apart from $2^κ≤|\mathcal P(\mathbb S)|$, we cannot say anything about this inequality in ZFC. If the generalized continuum hypothesis holds (or, less restrictively, if the powerset operator is injective on cardinals), then $|\mathcal G (\Bbb S)|<\mathcal |\mathcal P(\mathbb S)|$ for all successor cardinals $|\mathbb S|$, but by Easton's theorem it possible that $|\mathcal G (\Bbb S)| = \mathcal |\mathcal P(\mathbb S)|$ for (almost) all successors $|\mathbb S|$.

I do not believe there is much to say about $|\mathbb S|$ limit cardinal unless it is singular (see next section)

$|\mathcal G (\Bbb S)| = |\mathbb S|$

ZFC can prove that for singular $|\mathbb S|$ we have $|\mathcal G (\Bbb S)| > |\mathbb S|$.

But assuming the GCH, we do get partial positive result: $|\mathcal G (\Bbb S)|=|\mathbb S|$ for all successor cardinals $|\mathbb S|$

---

As a side note, $\mathcal G(A)$ is usually denoted as $[A]^{<|A|}$, and under the axiom of choice, $\mathcal I(A)$ is sometimes denoted as $[A]^{≥ω}$