one more question with cardinality: $(\{1,2,3\}^{\mathbb{N}} - \{1,2\}^{\mathbb{N}})\cap\mathcal{P}(\mathbb{N}\times\mathbb{N}).$ How can I calculate the cardinality of
$$\left(\{1,2,3\}^{\mathbb{N}} - \{1,2\}^{\mathbb{N}}\right)\cap\mathcal{P}(\mathbb{N}\times\mathbb{N}).$$
where $A^B$ is the set of all functions $f\colon B\to A$. 
thanks
 A: So you have all the functions from $\mathbb{N} \to \{1,2,3\}$ that are not in $\mathbb{N} \to \{1,2\}$?  Each of these functions is a subset of P($\mathbb{N} \times \mathbb{N}$), so that part is not important.  Hint: you just need one 3 in the image to take it out of $\mathbb{N} \to \{1,2\}$
A: Since the set of all functions from $\mathbb{N}$ to $\{1,2,3\}$ has cardinality $2^{\aleph_0}$, you certainly know that you have at most $2^{\aleph_0}$ elements in your difference.
So to show that $\{1,2,3\}^{\mathbb{N}} - \{1,2\}^{\mathbb{N}}$ has cardinality exactly $2^{\aleph_0}$, it is enough to produce $2^{\aleph_0}$ functions from $\mathbb{N}$ to $\{1,2,3\}$ whose image is not contained in $\{1,2\}$. 
Now, hopefully you agree that there are just as many functions from $\mathbb{N}$ to $\{1,2\}$ as there are from $\mathbb{N}-\{1\}$ to $\{1,2\}$? Now, take a function from $\mathbb{N}-\{1\}$ to $\{1,2\}$, and extend it to all of $\mathbb{N}$ in such a way that you guarantee the image includes $3$. That will produce at least $2^{\aleph_0}$ functions in $\{1,2,3\}^{\mathbb{N}}$ that are not in $\{1,2\}^{\mathbb{N}}$. 
