A question involving decimal expansions

Question

Let $E_k$ denote those numbers in $[0, 1]$ whose decimal expansions do not contain the digit 5 in the first $k$ decimal places.

Can someone explain me why $E_k$ may be regarded as a union of $9^k$ intervals of lengths $10^{-k}$?

Thank you!

Answer 1

First of all, there are $9^k$ sequences of $k$ digits from $\{0,1,2,3,4,6,7,8,9\}$, which we can assemble into $9^k$ elements of $[0,1)$ which just have $k$ digits after the decimal, none of them $5$. Call these numbers $e_0, e_1, \dots, e_{9^k-1}$, with $$e_0=0, e_1 = 0.\underbrace{00\cdots01}_k, \dots, e_{9^k-1} = 0.\underbrace{99\cdots99}_k$$ Then I claim that $$E_k = \bigcup_{0 \le i < 9^k} [e_i, e_i + 10^{-k}].$$ You should check that

• Every element of $[e_i, e_i + 10^{-k}]$ is an element of $E_k$. (This is true for the other endpoint $e_i + 10^{-k}$ only if we're allowed to write it with infinitely many trailing nines. E.g., is $0.5 \in E_k$ because $0.5 = 0.4999\dots$?)
• Every $x \in E_k$ is in some interval $[e_i, e_i + 10^{-k}]$. (What is $e_i$, in terms of $x$?)