Can the Cartesian Product of a set be in the power set of the original set? Consider a nonempty set $A$. Is it possible for $A\times A \subseteq A$?
I'm fairly certain that this can't be done for a finite set $A$, but what about a set with infinitely many elements? I have a hunch that one could say $a_i, a_j\in A$ implies that $(a_i , a_j)$ would need to also be in $A$, although then so would a Cartesian product with $(a_i,a_j)$ in at least one of its entries. Does anyone know the overall answer to this?
Note: I know very little set theory.
 A: The answer is yes. There are many examples but I'd argue the best example in a number of ways is $V_\omega$.
$V_\omega$ is the $\omega$th level of the cumulative hierarchy:


*

*$V_0=\emptyset$,

*$V_\lambda=\bigcup_{\alpha<\lambda} V_\alpha$ for $\lambda$ limit, and

*$V_{\alpha+1}=\mathcal{P}(V_\alpha)$.
It consists of the hereditarily finite sets. You know what it means for a set to be finite; well, a set is hereditarily finite if


*

*it is finite;

*all its elements are finite;

*all its elements' elements are finite;

*and so on.
It's not hard to show that if $X$ and $Y$ are hereditarily finite, so is $(X, Y)$. Hence, $V_\omega\supseteq V_\omega\times V_\omega$.

More elementarily, you can think of the set $K$ defined as follows:


*

*Let $K_0$ be any fixed set to start with.

*Let $K_{n+1}=K_n\cup(K_n\times K_n)$.

*Let $K=\bigcup_{n\in\mathbb{N}} K_n$.
This is essentially the same picture as $V_\omega$, but it might be easier to visualize first.

That said, there is an example of a finite set satisfying your property! HINT: what's the most annoying set you can think of? :P
A: Noah gave a good answer, which is probably the desired answer, but let me add something to it.
There is no mathematical reason to choose Kuratowski's encoding of ordered pairs ($(x,y)=\{\{x\},\{x,y\}\}$, and of course there are reasons to choose it: e.g. simplicity), and we can practically choose any encoding. For example, we can explicitly require that $\varnothing$ codes the ordered pair $(\varnothing,\varnothing)$, in which case $\{\varnothing\}\times\{\varnothing\}=\{\varnothing\}$.
But nevertheless, starting with any set $X$ and defining by induction: $A_0=X$; $A_{n+1}=(A_n\times A_n)\cup A_n$; then taking $A=\bigcup\{A_n\mid n\in\Bbb N\}$ will give us a set $A$ such that $A\times A\subseteq A$. And depending on your choice of coding, perhaps even equals to it.
