Let $X = \{ 0, 1 \}^{\mathbb{N}}$ be the D12: Set of boolean standard sequences.

Then $X$ is an D1195: Uncountable set.

Result R4057
on D12: Set of boolean standard sequences

The set of boolean standard sequences is uncountable

Formulation 0

Let $X = \{ 0, 1 \}^{\mathbb{N}}$ be the D12: Set of boolean standard sequences.

Then $X$ is an D1195: Uncountable set.

Proofs

Let $X = \{ 0, 1 \}^{\mathbb{N}}$ be the D12: Set of boolean standard sequences.

Assume to the contrary that $X$ is countable, that is, there exists a bijection $x : \mathbb{N} \to X$. Let $z : \mathbb{N} \to \{ 0, 1 \}$ be the D5362: Boolean Cantor diagonal sequence with respect to $x$. Since $z$ is a sequence which takes on values in the set $\{ 0, 1 \}$, then $z$ belongs to $X$. However, result R4058: Boolean Cantor diagonal sequence is not a term in the sequence inducing it shows that $x_n \neq z$ for every $n \in \mathbb{N}$. That is, there is an element in $X$ to which no element in $\mathbb{N}$ is mapped by $x$. Hence, $x$ is not a surjection, which contradicts the assumption that it is a bijection. Therefore, $X$ is uncountable. $\square$