Exercise 1.6.9

Question

Using the various tools and techniques developed in the last two sections (including the exercises from Section 1.5), give a compelling argument showing that $P(\mathbf{N}) \sim \mathbf{R}$.

Ulisse Mini · Accepted Answer

Recall from Exercise 1.6.4 that if
  $$
  S=\left\{\left(a_{1}, a_{2}, a_{3}, \ldotsight): a_{n}=0 	ext { or } 1ight\}
  $$
then $S \sim \mathbf R$. Define $f : P(\mathbf N) ightarrow S$ as $f(x) = (a_1, a_2, \ldots)$ where $a_i = 1$ if $i \in x$ and $a_i = 0$ otherwise. $f$ is thus a one-to-one, onto map between $P(\mathbf N)$ and $S$, hence $P(\mathbf N) \sim S$. Since $\sim$ is an equivalence relation, $P(\mathbf N) \sim \mathbf R$.

Exercise 1.6.9

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Exercise 1.6.9

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