so my question is, how can you prove that ${\Bbb Z}$ x ${\Bbb R}$ is uncountable?
So far I have tried proving that there is an uncountable subset of ${\Bbb Z}$ x ${\Bbb R}$ without luck and I'm not really sure what I can do
Thanks for any tips you can give!
Let us consider the following : $A\times B$ is countable. This means all subsets of $A\times B$ are countable. In particular $\{0\}\times B$ and $A\times \{0\}$ are countable. But the projection on $B$ and $A$ respectively induce bijection from $\{0\}\times B\to B$ and from $A\times \{0\}\to A$. This proves that $A$ and $B$ are countable
Write $|X|$ to mean the cardinality of the set $X$. You want to show that $|\mathbb N|<|\mathbb Z\times\mathbb R|$. In fact, something stronger is true: $|\mathbb R|\leq|\mathbb Z\times\mathbb R|$.
In general, we show that $|X|\leq|Y|$ by finding an injective function from $X$ to $Y$. (This is one definition of $|X|\leq|Y|$.) Can you think of an injective function from $\mathbb R$ to $\mathbb Z\times\mathbb R$? That is, can you find a way to match every element of $\mathbb R$ up with an element of $\mathbb Z\times\mathbb R$ without overlapping matches?
If $\mathbb{Z}\times\mathbb{R}$ were countable, we could pick a surjection $\pi=(\pi_1, \pi_2)$: $\mathbb{N}\longrightarrow\mathbb{Z}\times\mathbb{R}$. But then $\pi_2$: $\mathbb{N}\longrightarrow\mathbb{R}$ would be a surjection, so $\mathbb{R}$ would be countable.
