For each i ∈ N, let f_i : N → {0, 1}. Let A = {f_i : i ∈ N} and E = {n ∈ N : f_n (n) = 0}. Does there exist a f ∈ A such that E = {n ∈ N : f(n) = 1}.
I know that this question is related to set of binary sequence , but how can i answer the question it ask while relating it to Cantor's diagonal theorem?
Hint: Here's a table to help you understand the definitions being given. $$ \begin{array}{c|ccccc} i & f_i(0) & f_i(1) & f_i(2) & f_i(3) & \cdots\\ \hline 0 & \color{red}0&1&0&0&\cdots\\ 1 & 1&\color{red}1&0&1\\ 2 & 0&0&\color{red}1&1\\ 3 & 1&0&1&\color{red}0\\ \vdots &\vdots &&&&\ddots \end{array} $$ For this selection of functions $f_i$, we see that $0,3 \in E$ and $1,2 \notin E$. Correspondingly, the function $f$ being described would satisfy $$ f(0) = 1, \quad f(1) = 0, \quad f(2) = 0, \quad f(3) = 1. $$