Exercise 4.6.7

Question

In Section 4.1 we constructed functions where the set of discontinuity was $\mathbf{R}$ (Dirichlet's function), $\mathbf{R} \backslash\{0\}$ (modified Dirichlet function), and $\mathbf{Q}$ (Thomae's function).

\begin{enumerate}[label=(\alph*)] 
  \item Show that in each of the above cases we get an $F_{\sigma}$ set as the set where the function is discontinuous.
  \item Show that the two sets of discontinuity in Exercise $4.6 .1$ are $F_{\sigma}$ sets.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $\mathbf{R}$ is closed, so it is in $F_\sigma$, $\mathbf{R} \setminus \{0\} = \bigcup_{n=1}^\infty \mathbf{R} \setminus (-1/n, 1/n)$ is in $F_\sigma$ since $\mathbf{R}\setminus (-1/n,1/n)$ is closed, and finally $\mathbf{Q}$ is in $F_\sigma$ since $\mathbf{Q} = \bigcup_{n=1}^\infty \{r_n\}$ (where $r_n$ enumerate $\mathbf{Q}$, all countable sets are $F_\sigma$ sets.)
  \item Recall countable unions of $F_\sigma$ sets are $F_\sigma$ (see 3.5.2) and that open intervals are $F_\sigma$ sets, meaning $\mathbf{Z}^c = \bigcup_{z\in\mathbf{Z}} (z,z+1)$ is an $F_\sigma$ set. As for $\{x : 0 < x \le 1\} = (0, 1]$ I refer you to 3.5.3 (b).
   \end{enumerate}

Exercise 4.6.7

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Exercise 4.6.7

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