Show set of all real number in ($0,1$) with base $10$ decimal expansion contains no $3$s or $7$s is uncountable

Question

here is the question:

Show set of all real number in ($0,1$) with base $10$ decimal expansion contains no $3$s or $7$s is uncountable

My thoughts: to show it's uncountable, we should map it to an uncountable set like ($0,1$) in real numbers. And since we are looking at numbers with no $3$s or $7$s, we can consider a representation with only $\{0,1,2,4,5,6,8, 9\}$ and show this representation is a subset of ($0,1$). But right now I can't find such function and also I'm not sure that if we see a $3$ or $7$ in the decimals, should we map it to something else or how do we deal with it?

Any help is greatly appreciated. Thank you in advance.

Answer 1

Let $(0,1)_8$ be the set of all real numbers on the interval $(0,1)$ expressed in base 8. Let $S$ be the set you seek to show is uncountable, and let $s$ be an arbitrary element of $S$. We can take its decimal expansion, and replace each $9$ we find with a $7$, and each $8$ with a $3$. This forms a bijective (you can show this) mapping between $S$ and $(0,1)_8$. It's also obvious that $(0,1)_8$ is uncountable, so we are done.

Answer 2

It's well known that the set of binary sequences is uncountable. But, regarded as the decimal expresion of the fractional part of a real number in $[0,1]$ this set is a subset of your set. Then , you're done.

Answer 3

The representation of real number in $(0,1)$ with base 10 which contains 3 or 7 is of the form $\sum_{n=1}^{\infty}\frac{a_n}{10^n}$, where $a_n\in \{3,7\}.$

$\quad$ Since each $a_n$ has two choice, so the lebesgue measure of the above set is $\frac{2}{10}+\frac{2}{10^2}+...=\frac{2}{9}$. Therefore the set can't be countable, as Lebesgue measure of a countable set is 0. Hence the set is uncountable.