How do I find the cardinality of set with all the infinite binary sequences which consist of infinite "ones" and "zeros"? Set $S$ contains all the infinite binary sequences which must consist of infinite "ones" and infinite "zeros", first of all I am not sure if it is countable or not.. I thought about cantors diagonal but I am not sure about that, hope someone could help me here.
thank you in advance.
 A: Let $U = $ the set of all infinite series
$S = $ the set of all infinite series with infinite 1s and 0s
$A = $ the set of all infinite series with finite 0s
$B = $ the set of all infinite series with finite 1s.
$U = S \cup A \cup B$.
Claim 1: U is uncountable.
Claim 2: A and B are countable
Claim 3: the countable union of countable set is countable.
Hence, if $S$ were countable, $U$ would be countable.  It isn't so $S$ is uncountable.
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Claim 1: This is Cantor's diagonal.  If U were countable we could list all sequences so $U = \{A_i\} = \{\{a_{i_k}\}\}$.  Then the sequence $B = \{1-a_{i_i}\} \not \in U$.  A contradiction so $U$ is uncountable.
Claim 2: If an infinite sequence, {$a_n$} contains a finite number of 0s then there is an $a_m$ that is the last 0 value and all the remaining values are 1.  Consider the finite sequence {$b_n$} that has $m$ terms were $b_i = a_i$.  For every infinite sequence with finite number of 0 there is such a corresponding finite sequence.  And for every finite sequence we can find a corresponding infinite sequence by "sticking" an infinite number of 1s on the end.
So there is a 1-1 correspondence between finite sequences  and infinite sequences with finite 0s.  
But the set of finite sequences is countable.  ($t(\{b_n\})=\sum_{i=0}^m b_n*2^i \in \mathbb N$ is a 1-1 mapping from the finite sequences to the natural numbers.)
So there are countably many infinite sequences with finite 0s, and likewise there are countably many infinite sequences with finite 1.
Claim 3: Let $\{S_i\}$ be a class of countable sets so that each  $S_j = \{s_{j,k}\}$ so $\cup_i S_i = \{s_{j,k}\}$ with some possibility of multiple representation. It suffices to show that $\mathbb N \times \mathbb N$ is countable.  Basically the map (0,0)-> 0, (1,0) ->1, (0,1) ->2, (2,0)->3, (1,1) -> 4, (0,2)->5 etc or $(i,j) \rightarrow \sum_{l=0}^{j+k}l + k$ is a 1-1 mapping. 
A: Let $S$ be the set of all binary sequences other than those with a finite number of ones or a finite number of zeros, and let $T$ be the set of all binary sequences minus $S$.  The question is whether $S$ is countable. 
Then $T$ is countable. Indeed, $T$ is the union of $U$ and $V$ where $U$ is the set of binary sequences with a finite number of ones, and $V$ is the set of binary sequences with a finite number of zeros.
To show that $U$ is countable, consider the counting function mapping each binary sequence $\{x_m\}$ with no $1$'s past position $k$ to the  (finite) number $n = \sum_{n=0}^{k} x_n2^n$.  (The usual diagonal proof that a counting function can't work does not say anything in this case, since we are not allowed to construct a sequence with arbitrarily late ones:  Such a sequence must have infinite ones, hence is not in $U$.)
Since $U$ is countable, $T$ is countable, and $S = P(\omega) - T$ is uncountable.  $S$ is 1-1 with $P(\omega)$, not $\omega$. 
