# Discontinuity set of function closed and $F_\sigma$

Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, $f:X\to Y$ 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 $x\in X$.] I want to prove that $D_\epsilon(f)$ is closed in $X$. And also that $D(f)$, the discontinuity set of $f$, is an $F_\sigma$ set in $X$.

[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 do not know how to approach the first part and I'd appreciate some advice. I was thinking about taking a Cauchy sequence in $D_\epsilon(f)$ and showing that it converges in $D_\epsilon(f)$. But probably there is a more suitable approach. (Maybe showing the complement is open? The complement would be the set of all points of continuity of $f$, for any $\epsilon>0$.

For the second part, we can deduce that $D(f)$ is $F_\sigma$ because $D(f)=\bigcup\limits_{\epsilon > 0} D_\epsilon(f)$. The only problem that I see here is that this union is uncountable. Is this permissible?

• The second part can be ameliorated by noting $$D(f) = \bigcup_{n} D_{1/n}(f).$$ Nov 30, 2017 at 8:37
• A $\sigma$-algebra is an algebra that is closed under countable unions. So per your (@sequence) own observation, you will want to go with the suggestion above. Nov 30, 2017 at 12:02

Suppose $p$ is a limit point of $D_{\varepsilon}\left( f\right)$, and let $\delta>0$ be given. Then we may find $x \in D_{\varepsilon}\left( f\right)$ such that $x \in B_X(p, \delta) \setminus \{p\}$ (definition of a Limit Point). Since $B_X(p, \delta) \setminus \{p\}$ is an open set in $(X, d_X)$, we may find $\rho>0$ so that $B_X(x,\rho) \subseteq B_X(p, \delta) \setminus \{p\}$. Since $x \in D_{\varepsilon}\left( f\right)$, we may find $y, z \in B_X(x,\rho)$ such that $d_Y(f(y), f(z))\ge \varepsilon$. And so we have shown that there exist $y,z \in B_X(p, \delta)$ such that $d_Y(f(y), f(z))\ge \varepsilon$. Therefore $p \in D_{\varepsilon}\left( f\right)$.