Prove that the set of discontinuities of a derivative of an everywhere differentiable function $f(x)$ is of 1st category. $$$$Let $f'(x)$ be a derivative of an everywhere differentiable function $f(x)$. Now as the set of discontinuities of any arbitrary functions can be written as a countable union of closed sets. So let $A$ be the set of discontinuities of $f'(x)$, then we can write $$A=\bigcup_{n=1}^{\infty}A_n$$ where all the $A_n$ are closed set. Now suppose that for $n=n_0$ the set $A_{n_0}$ is not nowhere dense then there exists an open interval $(p, q)$ such that for any interval $I$ in that open interval $(p, q)$ we have $$I \cap A_{n_0} \neq \phi$$ and hence $A_{n_0}$ is dense in the open interval $(p, q)$ and as $A_{n_0}$ is closed so it contains the interval $(p, q)$ and hence $f'(x)$ is entirely discontinuous on the open interval $(p, q)$, but as the derivative of an everywhere differentiable function cannot be entirely discontinuous on an interval, so a contradiction. $$$$Is My Proof Correct??
Asked
Active
Viewed 48 times
4
-
Two issues I see: (1) Regarding "we have $I \cap A_{n_0} \neq \phi$", I don't know how this directly follows from not being nowhere dense. Not being nowhere dense means being "somewhere dense", or dense in some interval (see here), and the intermediate step $I \cap A_{n_0} \neq \phi$ doesn't arise. (2) Are you allowed to assume that a derivative cannot be everywhere discontinuous? To me this is at least as nontrivial as everything else up to this point. If you can assume this, then I believe your proof is OK, modulo my comments in (1). – Dave L. Renfro Aug 16 '20 at 09:09