Is $\{1, 2, 3\}\times \Bbb Z$ uncountable? $\Bbb Z$ being the set of integers.  
My understanding is that a set is uncountable if it's greater than the set of $\Bbb N$. 
Might it be that I'm misunderstanding the question, and misinterpreting the '$\times$' which I'm currently interpreting as 'intersection'.  
 A: That set is not uncountable. In fact $(a, b) \mapsto 3b+a-1$ is a bijection from your set to $\mathbb{Z}$!
Explanation: Two sets $A, B$ have the same cardinality if there exists a bijection between them. That is, if there is a function $f: A \to B$ so that each $b \in B$ corresponds with exactly one $a \in A$. Now, the function I have given maps from $\{1, 2, 3\} \times \mathbb{Z} \to \mathbb{Z}$. It takes some element of the first set like $(2,7)$ and maps it to $3 \cdot 7+2-1=22$. You can check that each integer corresponds with exactly one ordered pair in $\{1, 2, 3\} \times \mathbb{Z}$. Thus this is a bijection! Thus, your set has the same cardinality as the integers. But the integers are in bijection with the naturals (see if you can find such a function) thus, all these sets have the same cardinality as the natural numbers. We refer to such sets as "countable".
A: The symbol $\times$ is not an intersection. Intersection is almost always represented by $\cap$. The $\times$ is a Cartesian product, so the set 
$$A\times B = \{(a,b) ~~| ~~a\in A, b\in B\}$$
Which means that $\{1,2,3\}\times\Bbb{Z}$ is basically $3$ "copies" of $\Bbb{Z}$, or equivalently a countably infinite number of copies of $\{1,2,3\}$. This can be put into bijection with the integers explicitly, as Alexander pointed out, or simply by thinking of a picture similar to the diagonalization argument proving that $\Bbb{Q}$ is countable, but going horizontally and in both directions. Therefore it is countable. 
