$E$ is nowhere dense

Question

Let $E$ be the set of all $x\in [0.4,0.777\dots]$ whose decimal expansion contains only digits $4$ and $7$. How can we show that $E$ is nowhere dense in $[0.4,0.777\dots]$? That is, there is no interval $(a,b)$ in $E$.

My Try: $E$ does not contain intervals of the form $(0.4\dots41,0.4\dots3)$ and $(0.4\dots48,0.4\dots444)$ etc. But I can't find some 'form' of intervals that are not in $E$, such that every interval that lies in $[0.4,0.777\dots]$ lies in one of those intervals.

Answer 1

Suppose $(a,b)\subset E$. Pick $x\in(a,b)$. Then its decimal expansion only contains $4$'s and $7$'s If it contains infinitely many $4$'s, there's always one of them which you can replace by a $5$ without leavinf $(a,b)$, which is impossible, since $(a,b)\subset E$. And if it has infinitely many $7$'s, you can always replace one of them by a $6$ without leaving $(a,b)$.

Answer 2

Lets say there is interval (a,b).

Then all numbers between $a$ and $b$ must be in your set.

Lets take decimal expansion of $a$ and $b$ and find first digit where they differ. Put for example $5$ there. You have a number that is between them, but is not in your set.

Therefore there can't be any interval.

Answer 3

Let $a< b$ and let $c = (2 a + b)/3$ and $d = (a + 2 b) /3$. Both numbers $\in (a, b)$, and $c<d$. If both $c$ and $d$ are in $E$, the first decimal where they differ has value $4$ in $c$ and $7$ in $d$. The number $e\in (c, d)$ obtained by replacing this decimal by $5$ in $c$ satisfies $e\not\in E$.