Closure of set of numbers whose ternary expansion have only $0,1$ in first$ $N places

Question

I am trying to show that $A_N$ is closed, where $A_N \subseteq [0,1]$ is the set of real numbers whose ternary expansions have only $0$ or $1$ in the first $N$ places but can be $0,1$ or $2$ after that. I am allowed to assume that every real number in $[0, 1]$ has a ternary expansion, which is unique except for recurring $2s.$

This is what I have done so far:

Take $x \in [0,1]\setminus A_N$. Then x has at least one $2$ in the first $N$ places of its ternary expansion.

So $x= \frac{a_1}{3} + ... \frac{a_N}{3^N} + $$\sum_{i=N+1}^{\infty}$$\frac{a_i}{3^i}$ where $a_i=2$ for some $i \in \{1,...,N\}$

Now suppose, for a contradiction, that $y \in (x-\frac{1}{3^i},x+\frac{1}{3^i}) \cap A_N$.

So $y=\frac{b_1}{3} + ... \frac{b_N}{3^N} + $$\sum_{i=N+1}^{\infty}$$\frac{b_i}{3^i}$ where $b_i\in \{0,1\}$ for all $i \in\{1,...,N\}$

We know that $|x-y| < \frac{1}{3^i}$, but I'm not sure how to proceed from here $-$ any hints would be much appreciated.

Answer 1

You have:

• $A_1=\left[\frac03,\frac23\right]$;
• $A_2=\left[\frac0{3^2},\frac2{3^2}\right]\cup\left[\frac3{3^2},\frac{3+2}{3^2}\right]$;
• $A_3=\left[\frac0{3^3},\frac2{3^3}\right]\cup\left[\frac3{3^3},\frac{3+2}{3^3}\right]\cup\left[\frac{3^2}{3^3},\frac{3^2+2}{3^3}\right]\cup\left[\frac{3^2+3}{3^3},\frac{3^2+3+2}{3^3}\right]$

and so on… So, each $A_N$ is a finite union of closed intervals. Therefore, it is closed.