Munkres Chapter 1 Section 7 Exercise 8 Let $X$ denote the two element set $\{0,1\}$; let $X^\omega$ denote the set of all the binary sequences; and let $B$ denote the set of countable subsets of $X^\omega$. 
Then how to see if $X^\omega$ and $B$ have the same cardinality? 
 A: Let $b = \{ b_1, b_2, \dots \} \in B$ and write 
$$b_1 = (b_{11}, b_{12}, b_{13}, \dots)$$
$$b_2 = (b_{21}, b_{22}, b_{23}, \dots)$$
$$b_3 = (b_{31}, b_{32}, b_{33}, \dots)$$
$$\text{etc}$$
Now use the zig-zag pattern to form a new sequence :
$$s = (b_{11}, b_{21}, b_{12}, b_{13}, b_{22}, b_{31}, \dots)$$
Then the mapping $b \mapsto s$ is an injection of $B$ into $X^{\omega}$.
EDIT Following your comments, here is a more detailed description :
To see that $f : B \rightarrow X^{\omega}$ is an injection of $B$ into $X^{\omega}$, we must show that $f(b) = f(c) \Rightarrow b = c$.
Suppose that $f(b) = f(c) = (a_1, a_2, a_3, \dots)$.  Then, by definition of $f$ we have :
$$a_1 = b_{11} = c_{11}, \tag{a}$$
$$a_2 = b_{21} = c_{21}, \tag{b}$$
$$a_3 = b_{12} = c_{12}, \tag{c}$$
$$ \text{etc} $$
Thus, 
$$\begin{align} b & = \{b_1, b_2, \dots \}
\\ & = \{ (b_{11}, b_{12}, \dots), (b_{21}, \dots), \dots \}
\\ & = \{ (c_{11}, c_{12}, \dots), (c_{21}, \dots), \dots \} \tag{by (a),(b),(c)}
\\& = \{c_1, c_2, \dots \} 
\\& = c
\end{align}$$
The converse follows by the same (reverse) argument.
A: how about this case?
let B, C be elements in B(collection), which has countable subsets of x^omega.
let be B={ (1,0,1,0,1,0,1,0,1,0,.....),
       (0,1,0,1,0,1,0,1,...,) }

C={(1,0,0,1,1,0, ...),
(0,1,1,0,0...),
(1,0,0,1, ...)}
your representation identifying the elements in B(collenction),
i think that is not unique
your function gives us the same value of B, C(my example, B, C) so that the function is not injective.....
A: This sort of argument requires the axiom of choice to hold. But the place where we use the axiom of choice is relatively mild. Here are some helpful steps along the way:


*

*Show that $(X^\omega)^\omega$ and $X$ have the same cardinality.

*Find a surjection from $(X^\omega)^\omega$ onto $B$.

*Find an injection from $X^\omega$ into $B$.

*Look at the Cantor-Bernstein theorem.

*Have a beer to celebrate a proof well done. (The beer can be replaced by anything of celebratory nature, of course.)


The axiom of choice gets in the place where we exchange the surjection from the 2nd step to an injection in the other direction. The existence of such injection implies, for example, the existence of an injection from $\omega_1$ into the real numbers, something we know is impossible without a smidgen of choice.
