# Properties of discontinuity of the second kind

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.

## 1 Answer

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, $$m. We can find $$\{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$$.