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.
If you use this definition:

$$\begin{alignat}{2}
   0 &amp; {} = \{\} &amp;&amp; {} = \emptyset,\\
   1 &amp; {} = \{0\} &amp;&amp; {} = \{\emptyset\},\\
   2 &amp; {} = \{0,1\} &amp;&amp; {} = \{\emptyset,\{\emptyset\}\},\\
   3 &amp; {} = \{0,1,2\} &amp;&amp; {} = \{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\}
 \end{alignat}$$

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 $\mathbb{N}\cap\mathbb{N}^2 \neq \emptyset$.
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 $\Bbb N$.)