Can we construct a monotonic function $f : \mathbb{R} \to \mathbb{R}$ such that there is a dense set in some interval $(a,b)$ for which $f$ is discontinuous at all points in the dense set? What about a strictly monotonic function?
My intuition tells me that such a function is impossible.
Here is a rough sketch of an attempt at proving that such a function does not exist: we could suppose a function satisfies these conditions. Take an $\epsilon > 0$ and two points $x,y$ in this dense set such that $x However, after this point, I am stuck. Could we somehow partition $(x,y)$ into $n$ subintervals and conclude that there must be some point on the dense set that is continuous?
Answer
Such a function is possible.
Let $\Bbb Q=\{q_n:n\in\Bbb N\}$ be an enumeration of the rational numbers, and define
$$f:\Bbb R\to\Bbb R:x\mapsto\sum_{q_n\le x}\frac1{2^n}\;.\tag{1}$$
The series $\sum_{n\ge 0}\frac1{2^n}$ is absolutely convergent, so $(1)$ makes sense. If $x $$\lim_{x\to {q_n}^-}f(x)=\sum_{q_k Thus, $f$ is discontinuous on a set that is dense in $\Bbb R$ (and in every open interval of $\Bbb R$).
No comments:
Post a Comment