Does $\mathbb{N} \cap \mathbb{N}^2 = \varnothing$? Definitions. A cardinal $\mathfrak{a}$ is said to be finite if $\mathfrak{a} \ne \mathfrak{a} + 1$. A finite cardinal is also called a natural integer. The natural integers form a set, denoted by $\mathbb{N}$. The set $\mathbb{N}^2$ is the Cartesian product of $\mathbb{N}$ with itself. In other words, $\mathbb{N}^2$ is the set of ordered pairs 
$$(m,n) = \bigl\{\{m\},\{m,n\}\bigr\}$$
where $m$ and $n$ are natural integers.
 A: If you use this definition:
$$\begin{alignat}{2}
   0 & {} = \{\} && {} = \emptyset,\\
   1 & {} = \{0\} && {} = \{\emptyset\},\\
   2 & {} = \{0,1\} && {} = \{\emptyset,\{\emptyset\}\},\\
   3 & {} = \{0,1,2\} && {} = \{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}
 \end{alignat}$$
then every element of $\mathbb{N}$ is either $\emptyset$ or contains $\emptyset$. This is false for the elements of $\mathbb{N}^2$ using the standard definition of ordered pair:
$$(a, b) \triangleq \{\{a\}, \{a,b\}\}$$
since $\{a\}\neq\emptyset$ and $\{a, b\}\neq\emptyset$. So $\mathbb{N}\cap\mathbb{N}^2=\emptyset$.
If you use an alternative definition, like $(a, b) \triangleq \{a, \{a, b\}\}$, then we can have $a = b = \emptyset \Rightarrow (a, b) = \{\emptyset, \{\emptyset,\emptyset\}\} = \{\emptyset, \{\emptyset\}\} = 2 \neq \emptyset$. Under this definition of ordered pair, we have $\mathbb{N}\cap\mathbb{N}^2 \neq \emptyset$.
A: I think that the answer to this question is $\mathbb{N}\cap\mathbb{N}^2=\varnothing$.
$\mathbb{N}$ is a set of numbers: $\mathbb{N} = \{0,1,2,\ldots\}$.  $\mathbb{N}^2$ is a Cartesian cross product of elements of $\mathbb{N}$.  That is, an element of $\mathbb{N}^2$ has the form $(a,b)$ where $a,b\in\mathbb{N}$.  An ordered pair $(a,b)$ is defined as $\{\{a\},\{a,b\}\}$.  Since no element of $\mathbb{N}$ takes this form, the intersection must be $\varnothing$.
A: In comments you say you're defining cardinals using Scott's trick -- that is, a cardinal is the set of all sets with a certain cardinality of minimal rank.
Then suppose there's a Kuratowski pair $\{\{a\},\{a,b\}\}$ that is also a cardinal. If $a\ne b$ then $\{a\}$ and $\{a,b\}$ have different cardinalities, so they're not elements of the same cardinal. Therefore $a=b$ and our pair has the form $\{\{a\}\}$. Since the element $\{a\}$ has one element, we must be looking at the number $1$ if we're looking at a cardinal at all.
The smallest-rank singleton is $\{\varnothing\}$, and there are no other singletons that has this rank. So under Scott's trick, $1=\{\{\varnothing\}\}$. And this is indeed a Kuratowski pair -- but, alas! it is $\langle\varnothing,\varnothing\rangle$, and $\varnothing\notin\mathbb N$ when we're using Scott's trick.
So with this representation $\mathbb N\cap \mathbb N^2$ is still empty.

On the other hand, we might define the natural numbers as $$ \varnothing, \{\varnothing\}, \{\{\varnothing\}\}, \{\{\{\varnothing\}\}\},\ldots $$
like Zermelo originally did. (Though it is pretty difficult to imagine how this could be extended to a representation of infinite cardinals without obvious seams showing).
In that representation (but still with Kuratowski's pair definition) we have $ n+2 = \langle n, n\rangle$ for all $n$, and therefore $$\mathbb N\cap\mathbb N^2= \{2,3,4,5,\ldots\} $$
A: Note: No, you absolutely cannot take $$(\forall x)(\forall x^\prime)(\forall y)(\forall y^\prime)\big((x,y) = (x^\prime,y^\prime) \implies x = x^\prime \text{ and } y = y^\prime\big).$$ as the definition of ordered pairs! The above is a property of ordered  pairs, but it does not specify what $(x,y)$ is. In fact  there's a standard definition of $(x,y)$ in set theory.
Hint: If $x$ and $y$ are sets then $(x,y)\ne\emptyset$ and also every element of $(x,y)$ is nonempty.
(Of course the answer depends on the definition of $\Bbb N$.)
