Two claims about $L=\{a^n: n>o, \ n\text{ does not contain the digit } 7\}.$ that I didn't succeed to prove

Question

Let $$L=\{a^n: n>0, \ n\text{ does not contain the digit } 7\}.$$ (Examples: $n$ can be $12801$. $n$ can't be $71$)

Need to prove:

1. Let it be $w\in L$ and let it be $xyz$, a brakdown of w, so that $w=xyz$, $|y|>0$ ($x$ and $z$ can be the empty word $\epsilon$). Prove that there is an infinite number of $n$, so that $xy^nz∈L$.
2. Let it be $w\in L$ and let it be $xyz$, a brakdown of w, so that $w=xyz$, $|y|>0$ ($x$ and $z$ can be the empty word $\epsilon$). Prove that there is an infinite number of $n$, so that $xy^nz \not \in L$.

My problem here is that I don't know how to get the digit $7$ for $xy^nz$, (to prove claim 2), when the length of $y$ isn't known. That's why I also can't prove claim $1$.
How can I find $n$ for proving claim $1$ (or $2$), when I don't know anything about $|y|$?

Answer 1

To get the digit 7 for $xy^nz$ the easiest way is to get very large number.

For example if $|xyz|=104$ and $|y|=10$ you can find $n$ large enough to have $|xy^nz|=7004$ then for $n$ larger $|xy^nz|=70004$ then $|xy^nz|=700004$ and so on ...

The idea is to use large $n$ to get a $7$ in the high digits.

Proof:

Let $j$ be such that $10^j\leq |xyz| \leq 10^{j+1}$. Let $n_1=10^{j+1}+1$ and $n_2=10^{j+2}+1$. we know: $$|xy^{n_1}z|=|xyz|+|y^{10^{j+1}}|=10^{j+1}|y|+|xyz|$$ and $$|xy^{n_2}z|=|xyz|+|y^{10^{j+2}}|=10^{j+2}|y|+|xyz|$$ and since $|y|\leq 10^{j+1}$ we know that there exists $n$ with $n_1\leq n\leq n_2$ such that $|xy^nz|$ start with a 7.

Is it clearer?

For the first part.

Let $w\in L$ and $xyz$ be such that $|y|>0$ and $xyz=w$. let $j$ be such that $|w|\leq 10^j$. Two cases:

• assume that $|y|$ does not contain a 7. For all $n$ of the form $n=10^{j+k}+1$ with $k\in \mathbb{N}$ we have $|xy^nz|=|xyz|+|y|10^{j+k}$ that does not contain a 7.

• Assume that $|y|$ contains a 7. there exists $n_0$ such that $|y^{n_0}|$ does not contain 7 (This part I'm not sure how o prove it yet, but I'm pretty sure it's true). Now you can do the same as the first point but for $j$ such that $|xy^{n_0}z|\leq 10^j$ and for $n$ of the form $n_0*(10^{(j+k)})+1$