Exercise 3.5.6

Question

Show how the previous exercise implies that the set $\mathbf I$ of irrationals cannot be an $F_{\sigma}$ set, and $\mathbf{Q}$ cannot be a $G_{\delta}$ set.

Ulisse Mini · Accepted Answer

Recall from 3.5.3 that $\mathbf Q$ is an $F_\sigma$ set,
  suppose for contradiction that $\mathbf I$ were also an $F_\sigma$ set. Then we could write
  $$
  \mathbf{Q} = \bigcup_{n=1}^\infty F_n \quad	ext{and}\quad \mathbf{I} = \bigcup_{n=1}^\infty F_n'
  $$
  Each $F_n$ and $F_n'$ must contain no nonempty open intervals, since otherwise $F_n$ would contain irrationals and vise versa.
  Combine the countable unions by setting $	ilde F_{2n} = F_n$ and $	ilde F_{2n-1} = F_n'$ to get
  $$
  \mathbf{R} = \mathbf{Q} \cup \mathbf{I} = \bigcup_{n=1}^\infty 	ilde F_n
  $$

But in 3.5.5 we showed
  $$\mathbf{R} 
e \bigcup_{n=1}^\infty F_n$$
  which gives our desired contradiction, hence $\mathbf{I}$ is not an $F_\sigma$ set and $\mathbf{Q}$ is not a $G_{\delta}$ set (take complements).

Exercise 3.5.6

Answers

Comments

Exercise 3.5.6

Answers

Comments

Add answer