Is the below general approach to exercise 2.17 in Rudin’s PMA valid? It seems a bit different from other answers posted so not sure if completely off base or on the right track. Excuse typos. Thank you.

2.17. Let $E \subset [0,1]$ denote the set of all $x$ with only 4s and 7s in their decimal expansion. Is $E$ countable? Is $E$ dense in $[0,1]$? Is $E$ compact? Perfect?

Suppose $E \subset [0,1]$ is the set of all real numbers with only 4 and 7 in the decimal expansion. To construct E, consider the following. The endpoints of the interval $E_1=[\frac{4}{9},\frac{7}{9}]$ are the members of E with either all 4s and 7s (i.e. 0.444... and 0.777...) in the decimal expansions, respectively. Removing the middle $\frac{4}{15}$ of this interval, or retaining the outer $\frac{1}{30}$ closed interval on each side yields $E_2=[\frac{4}{9},\frac{43}{90}]\bigcup[\frac{67}{90},\frac{7}{9}]$. The endpoints of these two intervals consist of $0.444..., 0.477..., 0.744...,$ and $0.777...$. This process can be carried forward by retaining the outer $\frac{10^{-(n-1)}}{3}$, or removing the open middle interval of length $\frac{10^{-(n-2)}*4}{15}$, from the closed intervals in $E_{n-1}$, so that $E_n$ consists of the union of $2^{n-1}$ closed intervals of length $\frac{10^{-(n-1)}}{3}$ with the endpoints of these intervals consisting of all possible combinations of 4 and 7 of the decimal expansion up to the nth place and the remainder of the expansion consisting of either all 4s or all 7s. This construction implies $E=\bigcap E_n$. Then proof of the above properties in the question is then similar to the proof of the same for the Cantor set.


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