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??
Set Of Discontinuities Of A Derivative
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calculus