Is there, for every set $X$, a set $Y$ for which $|Y| < |X|$ but $|\mathcal{P}(Y)| \geq |X|$?

Question

As the title says, my question is:

Is there, for every set $X$, a set $Y$ for which $|Y| < |X|$ but $|\mathcal{P}(Y)| \geq |X|$?

I'm fairly certain this is true for finite sets but maybe false in the infinite case, for example if $X = \Bbb N$. There is no finite set for which its power set is equal in cardinality or larger than $\Bbb N$?

But I do not know if this indeed is the only counterexample, any assistance is appreciated.

Answer 1

No, not for every infinite set. Consider $\beth$ numbers:

1. $\beth_0=\aleph_0=|\Bbb N|$.
2. $\beth_{\alpha+1}=2^{\beth_\alpha}$.
3. For limit ordinals, $\beth_\delta=\sup\{\beth_\alpha\mid\alpha<\delta\}$.

Consider now $\beth_\omega$, where $\omega$ is the least infinite ordinal, that is $\beth_\omega=\sup\{\beth_n\mid n\in\Bbb N\}$, where $\beth_1=2^{\aleph_0}$ and $\beth_2=2^{2^{\aleph_0}}$ and so on.

If $|Y|<\beth_\omega$ then there is some $n$ such that $|Y|\leq\beth_n$, but then $|\mathcal P(Y)|\leq\beth_{n+1}<\beth_\omega$.

Answer 2

A cardinal number $\kappa$ such that $\lambda < \kappa$ implies $2^\lambda < \kappa$ is called a strong limit cardinal. $\aleph_0$ is one example of a strong limit cardinal, but it is not the only one.

Let $\kappa_0 = \aleph_0$, and define $\kappa_{n+1} = 2^{\kappa_n}$ for all $n \in \mathbb{N}$. Then $\kappa = \sup_{n \in \mathbb{N}} \kappa_n$ is also a strong limit cardinals (and is strictly greater than $\aleph_0$). (It is strong limit since for any cardinal $\lambda < \kappa$ there must be an $n \in \mathbb{N}$ such that $\lambda \leq \kappa_n$, and so $2^\lambda \leq 2^{\kappa_n} = \kappa_{n+1} < \kappa$.)

A similar construction starting with $\kappa_0$ an arbitrary cardinal will result in a strong limit cardinal greater than that cardinal.