P2776

Assume to the contrary that there is $n \in \mathbb{N}$ such that $x_n = z$. Then, in particular, $x_{n, n} = z_n$. But by definition of $z$, one has $z_n = 1 - x_{n, n}$. That is, by transitivity of equality, one has $x_{n, n} = 1 - x_{n, n}$. Adding $x_{n, n}$ to each side, we have $2 x_{n, n} = 1$ and thus $x_{n, n} = 1 / 2$. But $x$ was defined to be an element in $\{ 0, 1 \}^{\mathbb{N}}$, which necessitates that every term of $x_n$ takes a value in $\{ 0, 1 \}$, a set which does not contain $1/2$. This is a contradiction. Hence, there is no $n \in \mathbb{N}$ such that $x_n = z$. In other words, $x_n \neq z$ for all $n \in \mathbb{N}$. $\square$