Set of infinite binary sequence and Cantor's theorem

Question

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?

What do you mean by "Cantor's diagonal theorem"? There is something known as "Cantor's diagonal argument" and a result known as "Cantor's theorem", but there is no "Cantor's diagonal theorem". — Ben Grossmann, Nov 20 '20 at 15:29
ya ya it's cantor's theorem. sorry for the misleading question? — sipsip, Nov 20 '20 at 15:44
I just looked through the answer and I think it’s way outside out of my syllabus but anyways thanks for helping out... — sipsip, Nov 21 '20 at 06:07

Ben Grossmann · Accepted Answer · 2020-11-20T21:52:18.850

0

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 $$ f(0) = 1, \quad f(1) = 0, \quad f(2) = 0, \quad f(3) = 1. $$

edited Nov 20 '20 at 21:52

answered Nov 20 '20 at 15:46

Ben Grossmann

225,327

So to answer the question does there exist f in A, is yes right according to what you have shown? – sipsip Nov 20 '20 at 19:25
@sipsip The answer to the question is no. – Ben Grossmann Nov 20 '20 at 21:50
Emm don’t understand why you said no. – sipsip Nov 21 '20 at 05:46
The function constructed in this way could not possibly be an element of $A$ – Ben Grossmann Nov 21 '20 at 13:16
alright thank you for helping out – sipsip Nov 21 '20 at 17:12

Set of infinite binary sequence and Cantor's theorem

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