The set of discontinuous points is countable union of closed sets Im reading chapter9 Category, Real Analysis, Carothers, 1ed, talking about discontinuous functions of metric space. Here is a proof for a theorem that the set of discontinuous points is the countable union of closed sets in R:


Definition of D(f) is:

Definition of $ω_f(a)$ is:

Definition of ω(f; I) is:

Actually, I have some questions,:


*

*Let $F_n= \{a : ω_f(a) \geq 1/n \}$, then $F_1$ belongs to $F_2$, $F_2$ belongs to $F_3$, $F_3$ belongs to $F_4$ and so forth, is that right?

*What's the reason for "why"? I mean if question1 is yes, then union of $F_n$ seems to be no sense cos union of $F_n$ will = $lim F_n$ when $n$ goes to infinity.

*In $\{a : ω_f(a) \geq \xi \text{ for some } \xi > 0\}$, $\xi$ should as small as possible(basically, $\xi>0$), is that right?

*Is {a:$ω_f(a) < r$} a subset of or equivalent to ${I: ω(f:I) < r}$ in $\mathbb{R}$?
 A: *

*Yes: if $\omega_f(a)\ge\frac1n$, then certainly $\omega_f(a)\ge\frac1{n+1}$, so $F_n\subseteq F_{n+1}$ for each $n$.

*The statement that $D(f)=\bigcup_{n\ge 1}F_n$ really does require proof, though the proof is very easy. First off, it’s clear that if $\omega_f(a)\ge\frac1n$ then $\omega_f(a)>0$, so $F_n\subseteq D(f)$ for each $n\ge 1$. Thus, $D(f)\supseteq\bigcup_{n\ge 1}F_n$. Now suppose that $a\in D(f)$. Then $\omega_f(a)>0$, so there is some positive integer $k$ such that $\frac1k<\omega_f(a)$. But then $a\in F_k\subseteq\bigcup_{n\ge 1}F_n$, and since $a$ was any element of $D(f)$, this implies that $D(f)\subseteq\bigcup_{n\ge 1}F_n$. Combining the inclusions, we see that $D(f)=\bigcup_{n\ge 1}F_n$, as claimed.

*Do you mean in the equality $\{a:\omega_f(a)>0\}=\{a:\omega_f(a)\ge\epsilon\text{ for some }\epsilon>0\}$? That step could just as well be omitted: in (2) above I showed directly that $D(f)=\bigcup_{n\ge 1}F_n$ without using $\{a:\omega_f(a)\ge\epsilon\text{ for some }\epsilon>0\}$ at all. I don’t think that it contributes anything to one’s understanding of the argument.

*$\{a:\omega_f(a)<r\}$ is a set of real numbers, a subset of $\Bbb R$; $\{I:\omega(f;I)<r\}$ is a set of bounded intervals in $\Bbb R$. These two sets don’t even have the same kind of elements: the elements of the first set are individual real numbers, while the elements of the second set are bounded intervals of real numbers. Thus, the two sets are very different: not only is it impossible for them to be equal to each other, it’s impossible for either of them to be a subset of the other.
