Exercise 4.6.8

We do this by showing the complement is open. Let $x\in {\left(D_f^{\alpha }\right)}^c$, by the definition of $\alpha$-continuity there exists a $\delta >0$ such that $y,z\in V_{\delta }(x)$ have $|f(y)-f(z)|<\alpha$. To see openness notice $V_{\delta ∕2}(x)\subseteq {\left(D_f^{\alpha }\right)}^c$ since any $x^{\prime }\in V_{\delta ∕2}(x)$ is $\alpha$-continuous with ${\delta }^{\prime }=\delta ∕2$.