Verifying that $\mathcal{P}(\mathbb{N})\times\mathcal{P}(\mathbb{N})=\text{ the }\sigma\text{- algebra generated by }\mathbb{N}\times\mathbb{N}$ I am asked to show that the product $\sigma$-algebra of $\mathcal{P}(\mathbb{N})$ with $\mathcal{P}(\mathbb{N})$ consists of all subsets of $\mathbb{N}\times\mathbb{N}$, or in other words that $\mathcal{P}(\mathbb{N})\times\mathcal{P}(\mathbb{N})=\mathcal{P}(\mathbb{N}\times\mathbb{N})$ (where $\mathcal{P}(\mathbb{N}\times\mathbb{N})$ denotes the algebra generated by $\mathbb{N}\times\mathbb{N}$). The power set of the natural numbers, $\mathbb{N}$, is here denoted $\mathcal{P}(\mathbb{N})$. 

I am unsure how to go about doing this. I feel like 
\begin{align}
\mathcal{P}(\mathbb{N})\times\mathcal{P}(\mathbb{N}) &amp;= \{A:A\subset\mathbb{N}\}\times\{B:B\subset\mathbb{N}\} \\ &amp;= \{(A,B):A\subset\mathbb{N},B\subset\mathbb{N}\} \\
&amp;= \{(A,B):A,B\subset\mathbb{N}\}
\end{align}
is exactly all possible subsets of $\mathbb{N}\times\mathbb{N}$, and as such should be just equal to $\mathcal{P}(\mathbb{N}\times\mathbb{N})$. However, I am given the hint that $\mathbb{N}\times\mathbb{N}$ is countable, which leads me to think there is more going on here. Both $\mathcal{P}(\mathbb{N})$ and $\mathcal{P}(\mathbb{N}\times\mathbb{N})$ are uncountable, so I don't know how to put this information to use.

Any hints would be greatly appreciated!

Edit: Attempted to make the notation more clear.
The product $\sigma$-algebra of $\mathcal{P}(\mathbb{N})$ with itself (usually denoted $\mathcal{P}(\mathbb{N}) \otimes \mathcal{P}(\mathbb{N})$) is a $\sigma$-algebra of $\mathbb{N} \times \mathbb{N}$, so it will automatically be a subset of $\mathcal{P}(\mathbb{N} \times \mathbb{N})$. You just need to show the reverse inclusion: any set in $\mathcal{P}(\mathbb{N} \times \mathbb{N})$ will also be in the product $\sigma$-algebra.

Recall that by definition, $\mathcal{P}(\mathbb{N}) \otimes \mathcal{P}(\mathbb{N})$ is the $\sigma$-algebra generated by sets of the form $A \times B\subset \mathbb{N} \times \mathbb{N}$ where $A,B \in \mathcal{P}(\mathbb{N})$. In particular $A$ and $B$ can be one-point sets, which shows that one-point sets $\{(a,b)\}$ are in the product $\sigma$-algebra. Now use the properties of $\sigma$-algebras to show that consequently, any set in $\mathcal{P}(\mathbb{N})$ is also in the product $\sigma$-algebra.
You have two injections:
$\mathcal{P}(\mathbb{N})\subset\mathcal{P}(\mathbb{N})\times\mathcal{P}(\mathbb{N})\subset\mathcal{P}(\mathbb{N}\times\mathbb{N})$.

On the other hand, $\mathbb{N}=\mathbb{N}\times\mathbb{N}$, hence $\mathcal{P}(\mathbb{N})=\mathcal{P}(\mathbb{N}\times\mathbb{N})$. 

QED.