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Using Rudin's definition of a discontinuity of the second kind for a function. f has a discontinuity of the second kind if either $f(x^+)$ or $f(x^-)$ does not exist.

Supposing that $f$ has a discontinuity of the second kind on an interval $(a, b)$. I want to show that there is subinterval $(c, d)\subseteq (a, b)$ such that $f$ is not monotonic on any interval $(e, f)\subseteq (c, d)$.

We already know that if $f$ is monotonic on any open interval, then it cannot have any discontinuities of the second kind.

I can't figure out how to construct the interval around the discontinuity. My attempt so far is

Assume without loss of generality that $f$ has a discontinuity of the second kind at $x\in (a, b)$ and that $f(x^+)$ does not exist. Then there must exist a sequence $\{t_n\}\in(x, b)$ such that $t_n\rightarrow x$ but $f(t_n)$ does not converge.

I draw a blank once I get here, this could very well be false.

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You are on the right track. Let $m=\liminf_{n\to\infty}f(t_n)$ and $M=\limsup_{n\to\infty}f(t_n)$. Since $f(t_n)$ does not converge, $m<M$. We can find $\{a_n\}$ and $\{b_b\}$ subsequences of $\{t_n\}$ and $\delta>0$ such that $\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=x$ and $$ f(a_n)-f(b_m)>\delta\quad\forall m,n. $$ Then $f$ is not monotonic on $(x,z)$ for some $z>x$.

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