Exercise 1.6.2

Question

%1.6.2
  Let $f : \mathbf{N} 	o \mathbf{R}$ be a way to list every real number (hence show $\mathbf R$ is countable).

Define a new number $x$ with digits $b_1b_2\ldots$ given by
  $$
  b_{n}= \begin{cases}2 & 	ext { if } a_{n n} 
eq 2 \ 3 & 	ext { if } a_{n n}=2\end{cases}
  $$

\begin{enumerate}[label=(\alph*)] 
  \item Explain why the real number $x=. b_{1} b_{2} b_{3} b_{4} \ldots$ cannot be $f(1)$.
  \item Now, explain why $x 
eq f(2)$, and in general why $x 
eq f(n)$ for any $n \in \mathbf{N}$.
  \item Point out the contradiction that arises from these observations and conclude that $(0,1)$ is uncountable.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item The first digit is different
  \item The nth digit is different
  \item Therefore $x$ is not in the list, since the nth digit is different
   \end{enumerate}

Exercise 1.6.2

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Exercise 1.6.2

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