Countable / Uncountable Sets Let ${A_n}$, n = 1, 2, 3, ... be a sequence of countable sets, and put $B = A_1 \times A_2 \times \cdots \times A_n \times \cdots$. Show that B is uncountable. Prove that the same statement holds if each $A_n$ = {0, 1}. 
I'm not exactly sure how I should begin this. Any help would be appreciated.
 A: Assuming that for you countable means in bijection with the set of natural numbers, as each $A_{n}$ is countable, $\vert A_{n}\vert=\vert\mathbb{N}\vert$. Therefore one has
$$\vert B\vert = \vert\prod_{n\in\mathbb{N}}A_{n}\vert = \vert\prod_{n\in\mathbb{N}}\mathbb{N}\vert =\vert\mathbb{N}^\mathbb{N}\vert\geq\vert 2^{\mathbb{N}}\vert =\vert\mathbb{R}\vert > \vert\mathbb{N}\vert ,$$
where the last equality is due (for example) to the binary representation of real numbers. This chain of inequalities proves both your statements.
A: Suppose $B$ is countable, then let $b_n$ enumerate $B$. Define $d$ as follows:
If $b_n(n) = a_n\in A_n$ then pick any $a'_n\in A$ not equal to $a_n$. Then $d(n) = a_n'$.
$d$ must be an element of $B$ as it's $n$th componant is an element of $A_n$, however $d\neq b_n$ for each $n$, as $d$ disagrees with $d_n$ on the $n$th place. Hence $d$ is not in the enumeration of $B$. A contradiction.
Hence $B$ is uncoutable. 
A: $\{0,1\}\times\{0,1\}\times\cdots=\{0,1\}^\mathbb N=2^\mathbb N$ is not countable because no function $f:\mathbb N\rightarrow2^\mathbb N$ is surjective because no $n\in\mathbb N$ maps to $M=\{k\in\mathbb N|k\notin f(k)\}$.
$f(n)=M \Rightarrow (n\in M\Leftrightarrow n\notin M)$.
