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I am trying to prove that, given an arbitrary function $f$, the set of discontinuities of $f$, denoted $D_f=[{x\in{\Bbb{R}}:\mbox{f is not continuous at $x$}}$], is a $F_\sigma$ set - that is, a countable union of closed sets. This question may have been asked before, but after searching thoroughly, this is the only one to approach it in a certain manner with specific goals (taken from Stephen Abbott's Understanding Analysis).

First, a crucial piece of terminology (for those who do not know it): "Let $f$ be defined on $\Bbb{R}$, and let $\alpha>0$. The function $f$ is said to be $\alpha$-continuous at $x\in{\Bbb{R}}$ if there exists a $\delta>0$ such that for all $y,z\in{V_\delta{(x)}}$ it follows that $|f(y)-f(z)|<\alpha$. Also, we let $D_\alpha=[x\in{\Bbb{R}}:\mbox{f is not $\alpha$-continuous at $x$}]$

There are 4 steps in this proof, 3 of which I have already proved:

1) Show that, for a fixed $\alpha>0$, $D_\alpha$ is closed (already proved myself)

2) Show that, if $\alpha_1<\alpha_2$, then $D_{\alpha_2}\subset{D_{\alpha_1}}$ (already proved myself)

3) Let $\alpha>0$ be given. Show that if $f$ is continuous at $x$, then it is also $\alpha$-continuous at $x$. Show therefore that $D_\alpha\subset{D_f}$ (already proved myself)

4) Show that if $f$ is not continuous at $x$, then it is not $\alpha$-continuous for some $\alpha>0$. Show why this guarantees that $D_f=\bigcup_{n=1}^{\infty}D_{1/n}$. As, by point 1, each $D_{1/n}$ is closed, it thus follows that $D_f$ is a $F_\sigma$ set.

Now, I have proved the first sentence of point 4 myself, however, the bit where I run into trouble is how to show that this guarantees that $D_f=\bigcup_{n=1}^{\infty}D_{1/n}$. My idea was that this is true because continuity and $\alpha$-continuity are equivalent, so we simply take the union over all $\alpha$, and by point 2, the only $\alpha$ that 'matter' are those infinitesimally close to 0, and so as (1/n) converges to 0, all possible $\alpha$ infinitesimally close to 0 are 'covered'. This is a very loose argument that I would like to make formal, however, this only allows for $\alpha$ of the form $1/n$ even though $\alpha$ can be an irrational number... Please could someone try and explain how this is guaranteed thus (out of this v. long proof only bit I don't get bloody hell).

Thank you.

EDIT: $f$ is a function from the reals to the reals (although this argument can easily be extended to general complete metric spaces)

Pianoman1234
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  • Based on your description, $f: \mathbb{R} \rightarrow \mathbb{R} $ ? – ertl Aug 01 '18 at 22:56
  • @erti Yes, sorry for not making that clear, I'll edit that in now – Pianoman1234 Aug 01 '18 at 22:58
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    If $x \in D_f$ then $x \in D_\alpha$ for some $\alpha>0$ and hence by choosing ${1 \over n} \le \alpha$ we have $x \in D_{1 \over n}$, hence $x $ is in the union. – copper.hat Aug 01 '18 at 23:01
  • @copper.hat Oh yes seems rather trivial now, thanks for that. – Pianoman1234 Aug 01 '18 at 23:10
  • I'm not sure if step 1. is entirely correct. Consider the function $f : \Bbb R \to \Bbb R$ given by $$f(x) = \begin{cases}0 & x = 0, \ 2 & x \in \Bbb Q\setminus{0},\-2 & x \notin \Bbb Q.\end{cases}$$

    Then, $f$ is $3$-continuous at $0$ since $|f(x) - f(0)| \le 2$ for all $x \in \Bbb R$.

    On the other hand, $f$ is not $3$-continuous at any other point. Indeed, given any non-zero rational, there's always an irrational in any of its neighbourhood and vice-versa. The absolute difference of their function values is then $4$.

    Thus, $D_3 = \Bbb R \setminus {0}$, which is not closed.

     – Aryaman Maithani Feb 26 '21 at 04:58

1 Answers1

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If $0<b\leq a$ then $D_b\supset D_a.$ For any $a>0$ let $n(a)$ be the least (or any) $n\in \Bbb N$ such that $1/n\leq a.$ Hence $D_a\subset D_{n(a)}.$ Therefore $$D_f=\cup_{a>0}D_a\subset \cup_{a>0}D_{1/n(a)}\subset \cup_{m\in \Bbb N}D_{1/m}.$$ But since $\{1/m:m\in \Bbb N\}\subset \{a:a>0\}$ we also have $$\cup_{m\in \Bbb N}D_{1/m}\subset \cup_{a>0}D_a=D_f.$$

DanielWainfleet
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