Showing there exists no subintervals without digit. Let $A$ be the subset of $(0,1)$ containing the real numbers which have a 3 in their decimal expansion. 
As part of a larger result in an exercise in my text, I'm trying to show that $B = [0,1] \setminus A$ doesn't have any open interval $(x,y)$ with $0 \leq x < y \leq 1$. The prompt is to first show that for any $0 \leq x < y \leq 1$ there is a non-empty subinterval $(f,g) \subseteq (x,y)$ such that $(f,g) \subseteq A$. 
I'm having a hard time figuring out how to go about proving the prompt first of all. My first thought was to use the fact that between any two rationals/irrationals there is an irrational/rational and place $3$s on the end of decimal expansions to show that there must be some matching subset for that criteria, but I'm seriously doubting whether this approach is correct. 
 A: Consider the standard (i.e., we prefer things like $0.7000\ldots$ over 0.6999\ldots$ $) decimal digit representations $0.x_1x_2x_3\ldots$ and $0.y_1y_2y_3\ldots$ of $x$ and $y$. (The special case $y=1.0000\ldots$ is not covered by this, but not difficult).
As $x\ne y$, there is a first index $n$ such that the $x_n\ne y_n$, and as $x<y$, we must have $x_n<y_n$ for this $n$. As we use standard representations, there is $m>n$ such that $x_m\ne 9$.
Let $z$ be the number with representation $0.z_1z_2z_3\ldots$ where 
$$z_k=\begin{cases}x_k&k<m\\
x_m+1&k=m\\
3&k=m+1\\
0&k>m+1\end{cases}$$
Then


*

*$x<z$ because $z_m>x_m$ and $z_k=x_k$ for $k<m$,

*$z<y$ because $z_n<y_n$ and $z_k=y_k$ for $k<n$,

*and $z\in A$ because $z_{m+1}=3$.

A: Divide the interval $(0,1)$ into 10 subintervals of equal length: these sub-intervals will have end points of the for $k/10$ with $k=0,1,\ldots,10$.
All numbers of the fourth interval $(3/10,4/10)$ will have digit 3 as the first digit in the decimal expansion.
Now each of the 10 sub-intervals are further subdivided into 10 parts of equal length. In each of them consider the fourth subinterval  For example when $(7/10,8/10)$ is subdivided again fourth division would be $(73/10,74/10)$. The real numbers in each of these sub intervals will have 3 as the second digit in decimal expansion.
If we continue further in $n$th stage we will get intervals of length $1/10^k$.
Given an interval of length $\epsilon$ find $k$ such that $1/10^k$ is less than $\epsilon$. Then in $k$-th stage one of this sub interval will intersect the given interval.
