Prove that the points of discontinuity of an arbitrary function $\mathbb R \to \mathbb R$ are a $F_\sigma$ set.
Note: Proofs are available; this question is to verify and critique this proof and exposition.
Strategy: At first, it seems that points of discontinuity admit no restrictions: they can have any cardinality $\leq |\mathbb R|$, and can be individual points or intervals, open, closed, both, or neither.
Our strategy therefore is to introduce a relaxed version of continuity, called $\alpha$-continuous, which admits only open intervals, and use it to successively approximate continuity.
Definition: A function $f: A \subseteq \mathbb R \to \mathbb R$ is $\alpha$-continuous at $x \in A$ if for a particular fixed $\alpha > 0$ there exists a $\delta > 0$ such that for all $y, z$ such that $|y – x|, |z – x| < \delta \implies |f(y) – f(z)| < \alpha$.
Note that this definition differs from standard continuity in that it compares the image of two points in the neighorhood of $x$, as opposed to of $x$ itself and one point in its neighborhood. The reason for this change will become clear with the following lemma:
Lemma 1: For any $f: A \subseteq \mathbb R \to \mathbb R$, for any $\alpha > 0$, the points where $f$ is $\alpha$-continuous form an open set.
Proof: By its definition, if $f$ is $\alpha$-continuous at $x$, then $f$ is $\alpha$-continuous at some open interval surrounding $x$. The arbitrary union of open intervals is an open set.
Remark: As noted above, Lemma 1 is only true because we defined $\alpha$-continuous w.r.t any two points.
Lemma 2: For any $f: A \subseteq \mathbb R \to \mathbb R$, $f$ is continuous at $x \in A$ if and only if $f$ is $\alpha$-continuous at $x$ for all $\alpha > 0$.
Proof: If $f$ is continuous at $x$, then for any $\alpha > 0$, let $\varepsilon = \alpha/2$. All points in the $\delta$-neighborhood will then map within $(f(x) – \alpha/2, f(x) + \alpha/2)$ and thus be within $\alpha$ of each other, showing $f$ is $\alpha$-continuous at $x$.
Conversely, if $f$ is $\alpha$-continuous at $x$ for all $\alpha > 0$, then for any $\varepsilon > 0$, let $\alpha = \varepsilon$. All points in the $\delta$-neighborhood will then be within $\varepsilon$ of $x$, showing $f$ is continuous at $x$.
Main Proof: Let $S_n = \{x \in A: f$ is not $\alpha$-continuous at $x$ for $\alpha = 1/n\}$. By Lemma 1, $S_n$ is the complement of an open set and hence closed. Let $S = \bigcup_{n \in \mathbb N} S_n$. Clearly, if $f$ is not $\alpha$-continous at $x$ for $\alpha = k$, it is not $\alpha$-continous at $x$ for $\alpha > k$, and so by Lemma 2, $S$ is precisely the set of points where $f$ is not continuous. Since $S$ is a $F_\sigma$ set, the proof is complete.
Remark: We thus tame the uncountability of a set by building it via successive approximations, the analyst's tool of choice. Each approximation is a closed set, their union an illuminating example of $F_\sigma$. Although our standard tools to tame infinite sets, namely countability and openness, both fail to directly apply to the points of discontinuity, successive approximations lets us introduce them nonetheless. When dealing with an infinity too vast to contemplate, successively approximate it via something simpler.
Open questions: Are there other ways to characterize the points of discontinuity, or is every $F_\sigma$ set admissable? And are there questions that can be answered via successive approximations using uncountable collections of $F_\sigma$ sets, taking this approach one meta level higher?
Discussion: Is the proof correct, rigorous, and well-written? How can it be improved? Are the remarks accurate and helpful? Any feedback, positive or negative, is appreciated.
Best Answer
Your proof is the same as that one and is correct, rigorous, and well-written. See also this post and many others, linked or related.
Your note after the definition of $\alpha$-continuity and your remark after lemma 1 are not convincing because if $\forall y,z\in V_\delta(x)\quad|f(y)-f(z)|<\alpha,$ then in particular $\forall y\in V_\delta(x)\quad|f(y)-f(x)|<\alpha,$ and conversely, if $\forall y\in V_\delta(x)\quad|f(y)-f(x)|<\alpha,$ then $\forall y,z\in V_\delta(x)\quad|f(y)-f(z)|<2\alpha$ (this is the proof your lemma 2). To me, the true "difference from standard continuity" is that it is an intermediate notion with $\alpha$ fixed, so that (for $f$ at $x$) standard continuity is equivalent to $\alpha$-continuity for all $\alpha>0$ (this is the statement of your lemma 2).
For me, the shorter the better but it is a matter of taste, so I will not propose an alternative shortened version of yours. Two minor improvements could be to replace "$\alpha$-continuous for $\alpha=1/n$" by "$1/n$-continuous" (and similarly with "$\alpha$-continuous for $\alpha=k$"), and to choose earlier a notation for the set of points where $f$ is (or is not) $\alpha$-continuous, so as to use it in your lemmas and main proof.
The converse is true in $\Bbb R$, i.e. any $F_\sigma$ subset of $\Bbb R$ is the set of discontinuities of some function $f:\Bbb R\to\Bbb R.$
Your theme is related to Oscillation and Borel hierarchy.
Edit: An important point which I did not notice before: The domain of your $f$ should not be $A\subset\Bbb R,$ but $\Bbb R$ itself. Or if you insist on keeping $f:A\to\Bbb R,$ the result will be an $F_\sigma$-subset not of $\Bbb R$ but of $A$, equipped with the induced topology.