Is product of countably many copies of $\{0,1\}$ uncountable? Let $X$ denote the two point set $\{0,1\}$ and let $X_j=\{0,1\}\forall j=1,2\dots$ let $Y=\Pi_{j=1}^{\infty}X_j$, I need to determine whether each of the following are true or false:


*

*$Y$ is countable

*$|Y|$=|[0,1]|

*$\bigcup_{n=1}^{\infty}\Pi_{j=1}^{n}X_j$ is uncountable

*$Y$ is uncountable.
I guess $Y$ is uncountable (4), but I can not prove it.
 A: HINTS


*

*On a meta-level, if the first is true then the other three are trivially false.

*Every real in $[0,1]$ has a binary decimal expansion $0.b_1b_2b_3\ldots$

A: Every  $a \in Y$ is an infinite series of zeros and ones. Think of any $b \in P(\mathbb N)$ and try to find the connection between this two.
A: Proof that 3 is false.
List the elements of the union of finite cartesian products of {0,1} by increasing $n$ in layers.
Within layer list them by increasing size of the corresponding binary number:
0, 1 $(n=1)$
00, 01, 10, 11 $(n=2)$
000, 001, 010, 011, 100, 101, 110, 111 $(n=3)$
0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111 $(n=4)$
....
Now label the list using the natural numbers, first by layer, then moving to the next layer. 
The elements in the $n$-th layer will be labeled by $2$$n$-$1$ through $2$$n+1$-$2$.
This provides a bijection between N and all elements of the union of finite cartesian products of {0,1}, so the union of finite cartesian products of {0,1} is countable.
And I think it is done without the assistance of my favourite set theoretical bogeyman, Axiom of Choice.

P.S.
Regarding point 1 in the original question, is there any relation between the infinite countable product of {0,1} and w1 (the first uncountable ordinal)?
