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Let $(X,d_X)$ be a complete metric space and $(Y,d_Y)$ be a metric space, $f:X\to Y$ be a function. Given $\epsilon>0$, define

$$D_\epsilon(f)=\left\{x\in X: \forall\delta>0, \exists y, z\in B_X(x,\delta)\mbox{ such that } d_Y(f(y), f(z))\ge \epsilon\right\}$$

[That is, $D_\epsilon(f)$ is an $\epsilon$-discontinuity set for $f$, where $B_X(x,\delta)$ is an open ball centred at $x\in X$.]

It is known that $D_\epsilon(f)$ is closed in $X$ and that it is an $F_\sigma$ set.

[An $F_\sigma$ means that $D(f)$ can be expressed as $\bigcup\limits_{k=1}^\infty F_k$, where $F_k$ are closed sets.]

I want to show that $D(f)$ is either meagre (that is, it can be expressed as a countable union of nowhere dense subsets of itself) or that it contains a non-empty open set.

I think I'm missing some piece of knowledge which would allow me to relate a function defined on a Banach space to the fact that

$$D(f)=\bigcup_k D_{1/k}(f)$$

is a union of sets with empty interior. I would definitely appreciate some hints.

Moreover, it should follow that there cannot exist a function $f:\mathbb{R}\to \mathbb{R}$ such that $D(f) = \mathbb{R}\setminus \mathbb{Q}$ is the set of discontinuity for $f$, since $\mathbb{R}\setminus \mathbb{Q}$ does not contain a non-empty open set. But why can't $\mathbb{R}\setminus \mathbb{Q}$ be expressed as a countable union of nowhere dense subsets of itself?

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  • $\begingroup$ In the case Y =R, a set is the set of discontnuities of a function if and only if it is an $F_{\sigma}$. $\endgroup$ – Kabo Murphy Dec 1 '17 at 8:34
  • $\begingroup$ @KaviRamaMurthy Since $\mathbb{R}\setminus \mathbb{Q}$ is uncountable, is this precisely the reason it cannot be $F_\sigma$? $\endgroup$ – sequence Dec 1 '17 at 9:00
  • $\begingroup$ Closed sets are $F_\sigma$. So any uncountable closed set is $F_\sigma$. The reason $\mathbb{R}\setminus \mathbb{Q}$ can't be $F_\sigma$ is Baire Category. If $\mathbb{R}\setminus \mathbb{Q} = \bigcup A_n$, with $A_n$ closed and $A_n \subset \mathbb{R}\setminus \mathbb{Q}$, then each $A_n$ is nowhere dense in $\mathbb{R}$ (since it's closed with empty interior). This would say $\mathbb{R}\setminus \mathbb{Q}$ is first category in $\mathbb{R}$. Since $\mathbb{Q}$ is first category, we'd have that $\mathbb{R}$ is first category -- but it's a complete metric space. $\endgroup$ – David Bowman Dec 10 '17 at 4:17
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Let $D=\bigcup\limits_{k=1}^\infty F_k$ be a union of closed sets $F_k$ of a topological space $X$. If $D$ contains no non-empty open subset of $X$ then each $F_k$ contains no non-empty open subset of $X$, so it is nowhere dense. Thus $D$ is meager.

If $X$ is a complete metric space then $X$ is Baire by Baire Category theorem (see, for instance [HC, Th. 2.4]). In particular, there cannot exist a function $f:\mathbb{R}\to \mathbb{R}$ such that $D(f) = \mathbb{R}\setminus \mathbb{Q}$ is the set of discontinuity for $f$, since $\mathbb{R}\setminus \mathbb{Q}$ does not contain a non-empty open set.

But why can't $\mathbb{R}\setminus \mathbb{Q}$ be expressed as a countable union of nowhere dense subsets of itself?

The space $\mathbb{R}\setminus \mathbb{Q}$ is Baire by [HC, Prop. 1.23]. Moreover, the space $\mathbb{R}\setminus \mathbb{Q}$ is metrizable by a complete metric as $G_\delta$ subset of a completely metrizable space $\Bbb R$ by [Eng, Th. 4.3.22]

References

[Eng] Ryszard Engelking, General Topology, 2nd ed., Heldermann, Berlin, 1989.

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[HC] R. C. Haworth, R. C. McCoy, Baire spaces, Warszawa, Panstwowe Wydawnictwo Naukowe, 1977.

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