Every countable subset $B\subset X^{\omega}$ (i.e., every $B\in\mathcal{B}$ ) can be represented (not uniquely, but injectively) as a countable sequence $B=(b_{1},b_{2},...)$ of its elements, i.e. as a countable sequence of countably infinite sequences of $0$ ’s and $1$ ’s: $b_{i}=(0,0,1,0,1,...)$ . Hence, $B$ can be represented (not uniquely, but injectively) as a countable table (with a countable number of columns equal to the number of elements in $B$ , and countably infinite number of rows) by putting each $b_{i}$ as column $i$ of the table. If there are finite number of columns in the resulting table (a finite number of elements in $B$ ) then we add countably infinite number of columns having the same sequence $b$ that does not belong to $B$ . This way, as it is easy to see, we have constructed a unique countably infinite table for each countable set $B\subset X^{\omega}$ (unique, because we can easily reconstruct set $B$ given its table).

More formally, given a countable $B\subset X^{\omega}$ , we would represent it as a sequence of its elements $(b_{1},b_{2},\ldots)$ , and map it to $x\in(X^{\omega})^{\omega}$ such that, if $i$ is no greater than the number of elements in $B$ , then $(x_{i})_{j}=(b_{i})_{j}$ , otherwise, when there are $n$ elements in $B$ and $i>n$ , for $j\le n$ , $(x_{i})_{j}=1-(b_{j})_{j}$ , and for $j>n$ , $(x_{i})_{j}=0$ . Note that for $i>n\ge j$ , $x_{i}\neq x_{j}$ . It is easy to see that the mapping is injective.

In its turn, every countably infinite table can be injectively associated with a countable sequence of $0$ ’s and $1$ ’s. We can use a bijection $h:\mathbb{Z}_{+}\rightarrow\mathbb{Z}_{+}\times\mathbb{Z}_{+}$ to define the $i$ th element of the sequence as the element of the table having coordinates $h(i)$ . Then, different tables will map to different sequences.

Note again, that different countable sets $B$ cannot be mapped to the same table, and different tables cannot correspond to the same final sequence of $0$ ’s and $1$ ’s. So, we have an injection one way.

Here is an example. Let $B$ be the set of all sequences $b_{i}=(\underbrace{1,...,1}_{i},0,0,...)$ . Since $B$ is countable, it can be represented as a sequence of its elements (for example, $(b_{1},b_{2},\ldots)$ ). This representation is not unique, but different $B$ ’s cannot have the same representation. Then, the infinite table (matrix) and the injection from the table to a sequence of $0$ ’s and $1$ ’s can be constructed, for example, as shown in Figure 1↓.

Vice versa, every sequence in $X^{\omega}$ corresponds to a subset containing this one sequence only (singleton).

Having the two injections, we conclude, according to Exercise 6(b), that the two uncountable sets have the same cardinality.