The bound of the cardinalities of the sets that are subsets of their power set

Question

Consider the set $A_0=\{\emptyset\}$. The power set of $A_0$ is $P(A_0)=\{\emptyset,A_0\}$. Hence $A_0\subseteq P(A_0)$. I can also define inductively $A_n=A_{n-1}\cup \{\{\{...\{\emptyset\}...\}\}\}$. I guess I cannot use this construction to get an infinite set $A$ with $A\subseteq P(A)$. Is there an infinite set with this property? Is there a bound for the cardinalities of such sets?

Answer 1

There is no bound to the possible size of such sets. In fact, the von Neumann cumulative hierarchy is a proper class of such sets, with arbitarily large cardinalities, which is constructed in a similar way to how you did it (or, at least, attempted to do it). They are defined as follows.

• Let $V_0 = \varnothing$;
• If $V_{\alpha}$ has been defined, let $V_{\alpha+1} = \mathcal{P}(V_{\alpha})$;
• If $\lambda$ is a limit ordinal and $V_{\alpha}$ has been defined for all $\alpha < \lambda$, then define $V_{\lambda} = \bigcup_{\alpha < \lambda} V_{\alpha}$.

You can verify that $V_{\alpha} \subseteq \mathcal{P}(V_{\alpha}) = V_{\alpha+1}$ for all ordinals $\alpha$.

More generally, any transitive set $X$ satisfies $X \subseteq \mathcal{P}(X)$. What it means for $X$ to be transitive is that, for all $x \in X$, if $y \in x$ then $y \in X$. This is the same as saying that $x \in X \Rightarrow x \subseteq X$. But $x \subseteq X$ means the same thing as $x \in \mathcal{P}(X)$, so a set $X$ is transitive if and only if $X \subseteq \mathcal{P}(X)$.