Cardinality of the set of infinite binary sequences

Question

Let $B := \{ (x_n) \mid x_n \in \{0, 1\}, n \in \mathbb N \}$ then prove that $|B| = 2^{\aleph_0}$.

I know that the given set $B$ is uncountable. This can be deduced by proving that any countable subset of sequences of $B$ will be a proper subset. $B$ being countable would then give a contradiction.

To explicitly find out the cardinality of $B$, however, is what the problem demands. Will it be correct to say that since there are exactly $2$ choices ($0$ or $1$) for each term of any infinite binary sequence, whose cardinality is ${\aleph_0}$, so, the cardinality of $B$ is $2^{\aleph_0}$?

Answer 1

A binary sequence $(x_n)$ is just a function $x: \mathbb{N} \to \{0,1\}$. The $x_n$ is an alternative notation for $x(n)$.

In cardinal arithmetic $\kappa^\lambda$, for two cardinals $\kappa,\lambda$, is defined as the cardinal number of the set of all functions from a set of size $\lambda$ to a set of size $\kappa$.

So the size of your $B$ (all binary sequences) is, by this definition, $|\{0,1\}|^{|\mathbb{N}|} = 2^{\aleph_0}$

Answer 2

You're correct, but an even easier way to see so is that for each such binary sequence, one can construct a unique subset $S$ of $\mathbb{N}$ by including the number $n$ in $S$ iff the $n$th term of the sequence is 1. Then, the set of binary sequences is in bijection with the set of subsets of $\mathbb{N}$, which is the definition of $2^{\aleph_0}$.