Is it possible that $A\subseteq A\times B$ for some non empty sets $A,B$? I was wondering if there exist two non empty sets $A,B$ such that
$$A\subseteq A\times B.$$
I know that always exists a subset of $A\times B$ with the same cardinality of $A$, but i'm requesting here that $A$ is a subset of $A\times B$ without using any identification map. 
At first i thought that this was not possible because $A$ and $B\times A$ are two sets containing different kind of elements: the second contains pairs like $(a,b)$ with $a\in A, b\in B$; the first just single elements $a\in A$. Moreover, suppose $A\subseteq A\times B$ holds and take $a \in A$. Then $a=(a_1,b_1)$ for some $a_1 \in A, b_1\in B$. For the same reason $a_1=(a_2,b_2)$ and so $a=((a_2,b_2),b_1)$. Following this argument I got some sort of recursive infinite definition for $a$ that made me suspect something is wrong. However if I take $$A=\mathbb{N}^{\mathbb{N}} ;B=\mathbb{N}$$
is it true that $A=A\times B$ or I'm missing something? Moreover, if $A\subseteq A\times B$ can be true, are there other examples?
edit: I add another strange example: take $A=\bigcup_{i=1}^{\infty} \mathbb{N}^i $ and $B=\mathbb{N}$, then $A \times B \subset A$. This makes me think that maybe exists also an example for the other inclusion.
 A: The existence of sets $A$ and $B$ with $A\subseteq A\times B$ contradicts the axiom of regularity (or the axiom of foundation.)  The precise proof depends on which construction of $A\times B$ you prefer.  We will use the most common formulation, in which we represent the ordered pair $(a,b)$ by the set $\{\{a\},\{a,b\}\}$, but the proof easily adapts to other constructions.
Suppose $A\subseteq A\times B$.  Since the sets are nonempty, we can pick $a_0\in A$.  By assumption, we can write $a_0=(a_1,b_1)$ with $a_1\in A$ and $b_1\in B$, and continuing on we can write $a_i=(a_{i+1},b_{i+1})$.  Under our formulation of the cartesian product, this means $a_{i+1}\in \{a_{i+1}\}\in a_i$.
This gives an infinite descending sequence
$$ \cdots \in a_{i+1}\in \{a_{i+1}\}\in a_i \in \{a_i\}\in a_{i-1}\cdots \in a_0,$$
contradicting the axiom of foundation.
A: An element of $\Bbb N^{\Bbb N}$ is an infinite sequence of natural numbers.  Buy what do we mean by a sequence.  $n_0,n_1,\ldots n_k,\ldots$ is really nothing more than a function $f:\Bbb N\to\Bbb N$.  For example, $f(k)=n_k$.
So, $\Bbb N^{\Bbb N}=\{f:\Bbb N\to\Bbb N\}$.  So, what is an element of $A\times\Bbb N$?  It is a pair $(f,n)$. The are not elements of $A$.
