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!!
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 $F_{\sigma}$.