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. 

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 $z_{m+1}=3$.

