Respuesta :

Answer:

You can prove this important result as follows:

Step-by-step explanation:

Let [tex]A[/tex] be the set of all binary sequences, that is to say, [tex]\{0,1\}^{\mathbb{N}}[/tex]. Suppose that [tex]A[/tex] is a countable set. Then the elements of  [tex]A[/tex] can be ordered as a sequence [tex]\{s_{1},s_{2}, s_{3},...\}[/tex], where each [tex]s_{i}[/tex] is a binary sequence. The [tex]k\text{-th}[/tex] digit of each sequence is expressed by [tex]s_{n}(k)[/tex]. Define the sequence [tex]s[/tex] as follows:

[tex]s(k)=\begin{cases}1&\text{if}\,s_{k}(k)=0\\ 0 &\text{if}\,s_{k}(k)=1\end{cases}[/tex]

Note that [tex]s[/tex] differ from each [tex]s_{k}[/tex] in at least one digit. Then [tex]s\neq s_n[/tex] for all [tex]n\geq 1[/tex], then [tex]s\notin A[/tex]. This contradicts the fact that [tex]A[/tex] is the set of all binary sequences. Then [tex]A[/tex] must be a uncountable set.

ACCESS MORE