A discontinuous function satisfying Intermediate Values Theorem.

Question

For each $x\in[0, 10)$, write its decimal expansion $x = a_0.a_1a_2a_3 \dots a_n\dots$. Take into account the number $x'$ formed by the digits of odd rank of this decimal expansion; i. e., $x'= 0.a_1a_3a_5 \dots a_{2n-1} \dots$. We have two cases:

(1) $x'$ is not periodic

(2) $x'$ is periodic after the digit $a_{(2n_0-1)}$

Now let $f$ be the function from the half-open interval $[0, 10)$ to $\mathbb{R}$ defined by

$f(x) = \begin{cases} 0 & \text{in case (1) for } x'\\ a_{(2n_0)}.a_{2(n_0+1)} \dots a_{2(n_0+k)} \dots & \text{in case (2) for } x' \end{cases}$

(Note that in case (2), the integer part $[f(x)]= a_{(2n_0)}$ of $f(x)$ can be nonzero.)

QUESTION

Prove that for all $x_1, x_2$, the function $f$ takes on the open interval $(x_1, x_2)$ all values between $f(x_1)$ and $f(x_2)$ despite $f$ being discontinuous in all points of its domain.

Answer 1

There is an even stronger property than required: On any inhabited open interval $(x_1, x_2)$ the function f takes all values in the interval $[0,10)$. As both $f(x_1) $ and $f(x_2) \in [0,10)$, this property is stronger than your question.

First we pick a suitable subset of $(x_1, x_2)$ to make think nicer. There is some number $t = t_0.t_1t_2t_3…t_{2n+1}$ such that both $t$ and $t+10^{-(2n+1)}$ are in $(x_1, x_2)$ and $t_{2n+1}$ is neither $0$ nor $9$.

To get any desired value $y = b_0.b_1b_2b_3…$ just construct $x_y = t_0.t_1t_2…t_{2n+1}b_00b_10b_20…$. Note that we are in case $(2)$ with $n_0 = n$.

There are probbably some issues for numbers without a unique decimal expansion – but your definition already ignores those. (I made a comment to clarify this.)