I need help with this. The teacher defined $\mathbb{N}^*$ as $\mathbb{N}^* =\{1,2,...\}$.
I tried to express an interval in decimal expansion like:
$$a<r<a+\frac{1}{10^k}$$
Where $a\in \mathbb{Z}$ and $r=d_0 + \frac{d_1}{10}+\frac{d_2}{10^2}+ \cdots + \frac{d_k}{10^k}+\cdots + \frac{d_n}{10^n}$. And it's supposed to satisfy that $d_k=0$. But i got stuck because how can I guarantee that all rationals in that interval have it's k-th decimal equal zero? Even more, how can I define $a$.
The teacher gave us an answer but it seems weird to me; the interval he gave us is: $$\bigcup_{i=0}^{10^{k-1}}\left[\frac{10 i}{10^k}, \frac{10 i+1}{10^k}\right]$$
Anyone could help me?
It often helps to pick a particular $k$. Let's do $k=2$. Where are all the rationals in $[0,1]$ that have a $0$ in the $2^{nd}$ place? They are $[0.00,0.10),[0.10,0.11)\ldots[0.90,0.91)$ You can have anything in the first place and anything in the $3^{rd}$ and on places. As you stated the question, any one of these would be a good interval for $k=2$. The generalization to larger $k$ written in this form should be easy to see.
Your teacher's answer is not an interval, it is a union of intervals. I suspect it is the answer to a slightly different question, asking for a set that includes all the rationals in $[0,1]$ with a $0$ in the $k^{th}$ place. If you plug $k=2$ into the expression, $i=0$ corresponds to the first interval in my first paragraph, $i=1$ to the second, and $i=9$ to the last. I have two issues with your teacher's answer if it is intended to the question I suggested. The upper limit on $i$ should be $10^{k-1}-1$, so in my example the union stops at $i=9$. Second, each interval in the union should be open at the right end because the expression includes $0.01$ in the first interval. Your teacher may be thinking of expressing $0.01$ as $0.0099999\ldots$