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Let $f:\Bbb R\to \Bbb R$ and $D=\{x\in \Bbb R: f $ is discontinuous at $x\}$.

My problem is : Is it possible to have $D=\Bbb P$ where $\Bbb P$ is the set of irrationals in $\Bbb R$.

I know the answer is negative, but, how to prove it??

My attempt: First, I proved that $\Bbb P$ is not a countable union of closed sets in $\Bbb R$.Then, I read somewhere that $D$ is an $F_{\sigma}$ set (but don't know how to prove it).

If one could prove the second part, the problem is solved, but How to do it?? Thanks in advance!!

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Also, partial duplicate:… – Ravi Aug 16 '12 at 8:22
$D$ is $F_{\sigma}$ set(which can be wriiten as countable union of closed sets) but $\Bbb P$ can't be written as countable union of closed sets in $\Bbb R\implies D$ can't be $\Bbb P$ for any function $f:\Bbb R\to \Bbb R$. I don't know how to prove that $D$ is an $F_{\sigma}$ set (i read it somewhere) – Aang Aug 16 '12 at 8:28
For a proof that $D$ is an $F_\sigma$ see this answer to the question cited by Fortuon. – Brian M. Scott Aug 16 '12 at 8:29
For the proof that irrationals are not $F_\sigma$, see How to show that $\mathbb{Q}$ is not $G_\delta$ and Example of a Borel set that is neither $F_\sigma$ nor $G_\delta$. – Martin Sleziak Aug 16 '12 at 12:13
up vote 4 down vote accepted

Denote $$G_{k}:=\bigcup\{U\subset\mathbb{R}:U\,\,\mathrm{is}\,\,\mathrm{open}\,\,\mathrm{and}\,\,|f(x)-f(y)|<\frac{1}{k}\,\,\mathrm{for}\,\,\mathrm{all}\,\,x,y\in U\}$$ for all $k\in\mathbb{N}$. Show that $D^{c}=\bigcap_{k=1}^{\infty}G_{k}$, making $D^{c}$ (i.e. the continuity points of $f$) a $G_{\delta}$-set and thus $D$ a $F_{\sigma}$. If you need some help concerning the steps I may expand this answer or give hints in the comment section.

And by the way, irrationals is not $F_{\sigma}$ because rationals is not $G_{\delta}$. If rationals were $G_{\delta}$, then as a countable completely metrizable topological space ($G_{\delta}$ subsets of a complete metric space are completely metrizable) it has an isolation point by Baire category theorem. But since rationals have no isolation points, this is a contradiction. Hence rationals is not $G_{\delta}$ and thus irrationals is not $F_{\sigma}$.

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What is an isolation point? Do you mean isolated point? Anyway: it seems much easier to note that the irrationals $\mathbb{R}\smallsetminus \mathbb{Q} = \bigcap_{q \in \mathbb Q} \mathbb{R} \smallsetminus \{q\}$ are a dense $G_\delta$ and if the rationals were a $G_\delta$, Baire would tell us that $\mathbb{Q} \cap (\mathbb{R}\setminus \mathbb{Q})$ is dense. – t.b. Aug 16 '12 at 10:38
@t.b. Yeah, I meant an isolated point. And true, that also does the job. Thanks for providing an alternative way of seeing it. – T. Eskin Aug 16 '12 at 14:10
Somehow I managed to overlook the word countable in the argument of the second paragraph, that's why I was puzzled... A few further arguments can be found in the links of Martin's comment to the question. – t.b. Aug 16 '12 at 14:24
@t.b. Yeah, the countability is important :) Baire implies that any countable completely metrizable topological space has an isolated point. If not, you could write the whole space as a countable union of nowhere dense closed sets (singletons), making the whole space to have an empty interior, which is a contradiction. – T. Eskin Aug 16 '12 at 14:27

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