# Wimpy powerset function

Define the 'wimpy powerset function' $\mathcal{W} : \mathrm{Set} \rightarrow \mathrm{Set}$ by writing $$\mathcal{W}(B) = \{X \in \mathcal{P}(B) : |X| < |B|\}.$$

A few preliminary observations.

1. If $B$ is finite, then $|\mathcal{W}(B)| + 1 = |\mathcal{P}(B)|.$

2. If $B$ is countable (e.g. take $B=\mathbb{N}$), then $|\mathcal{W}(B)| = |B|.$

What else is known about $\mathcal{W}$? In particular:

• What can we say about $\mathcal{W}(\aleph_1)$ and $\mathcal{W}(\beth_1)$?
• Do there exist sets $B$ such that $|\mathcal{W}(B)| = |\mathcal{P}(B)|$?
• Nice name, nice question. Commented Jul 13, 2013 at 4:56
• Just be patient, sooner or later Asaf Karagila will answer your question with or without assuming ZFC ;-) Commented Jul 13, 2013 at 6:03
• A nice example without choice: Consider, say, Solovay's model, where the size $\mathfrak c$ of the reals is a successor of $\aleph_0$, but $\aleph_1\not\le\mathfrak c$. It is a theorem of Tarski that for any set $X$, the collection $\mathsf{WO}(X)$ of well-orderable subsets of $X$ has size strictly larger than the size of $X$. In this case, this means that $|\mathcal W(\mathbb R)|>\mathfrak c$. But one can check that in this case $|\mathcal P(\mathbb R)|>|\mathcal W(\mathbb R)|$. Commented Jul 13, 2013 at 6:18
• Another remark without choice: A Dedekind finite set is a set $X$ such that $|Y|<|X|$ for any proper subset $Y$ of $X$. It is consistent without choice that there are infinite Dedekind finite sets. If $X$ is infinite and Dedekind finite, then either $\mathcal P(X)$ of $\mathcal P^2(X)$ is Dedekind infinite (both cases are consistent). If $Y$ is $X$ or $\mathcal P(X)$, whichever is Dedekind finite and has Dedekind infinite power set, then $\mathcal W(Y)$ has the same size as $\mathcal P(Y)$. On the other hand, if $\mathcal P(X)$ is Dedekind finite, then $|\mathcal W(X)|<|\mathcal P(X)|$. Commented Jul 13, 2013 at 6:23
• It is fun to come up with names, but this already has two notations: $[\kappa]^{<\kappa},\mathcal P_\kappa(\kappa)$. Commented Jul 13, 2013 at 7:28

We're assuming ZFC, right?

$|\mathcal W(\omega_1)|=\beth_1$.

$\beth_1\le|\mathcal W(\beth_1)|\le\beth_2$;
if $2^{\aleph_0}=\aleph_1$, then $|\mathcal W(\beth_1)|=\beth_1$, but
if $2^{\aleph_0}=\aleph_2$ and $2^{\aleph_1}=2^{\aleph_2}=\aleph_3$, then $|\mathcal W(\beth_1)|=\beth_2$.

$|\mathcal W(\beth_{\omega})|=|\mathcal P(\beth_{\omega})|=\beth_{\omega+1}$.

• 1. Yes, assume ZFC. 2. If you could spell out a little how you came to these conclusions, that would be great!! Commented Jul 13, 2013 at 5:55
• @user18921 What you need to verify is that if $\kappa$, $\lambda$ are infinite, with $\kappa\ge\lambda$, then $[\kappa]^\lambda$, the collection of subsets of $\kappa$ of size exactly $\lambda$, has size exactly $\kappa^\lambda$. Commented Jul 13, 2013 at 6:04
• @user18921: Pursuing Andres’s comment: if $\kappa\ge\lambda$, then $|\kappa\times\lambda|=\kappa$, and it’s easy to find $\kappa^\lambda$ distinct subsets of $\kappa\times\lambda$, each of power $\lambda$, so $\left|[\kappa]^\lambda\right|\ge\kappa^\lambda$. The opposite inequality is even easier. Commented Jul 13, 2013 at 8:02
• @Brian: In fact $^\lambda\kappa\subseteq[\lambda\times\kappa]^\lambda$. Commented Jul 13, 2013 at 8:22
• I guess I missed something. Why the rollback? Commented Jul 14, 2013 at 0:54