mjhjmtu

As the title suggests, I am having some trouble understanding an exercise regarding Borel sets. In particular I am trying to

Show that the set of real numbers that have a decimal expansion with
the digit $5$ appearing infinitely often is a Borel set.

My question here is what is meant by the phrase "infinitely often" in this case? I have never seen this phrase before in a formal context. This attempt of mine is horrific so don't butcher me, but I tried starting out in the following way:

Let $mathcalA_5$ denote the set of real numbers that have a decimal expansion with the digit $5$ appearing infinitely often, i.e. every $ainmathcalA_5$ can be written as $$a=ldots+ 5cdot10^n+5cdot10^n-1+ldots+5cdot 10 +5+5cdot10^-1+ldots 5cdot10^-m-1+5cdot10^-m+ldots$$

Am I interpreting "infinitely often" correct here? If not, what is it trying to say?

The phrase means just this; if $$a_k 10^k + a_k-1 10^k-1 + dots + a_0 + a_-1 10^-1 + a_-2 10^-2 + cdots$$
is the decimal expansion of a real number, then the set
$$
n : a_n = 5, -infty < n leq k
$$
is an infinite set.

Pulled up from the comments below: to elaborate, the given condition means that there is some infinite subsequence of $5$'s in the sequence $( a_n )$.

Take care that it does not mean that all $a_n$'s are $5$, or that all $a_n$'s after a certain point are $5$. For example, you could have that $a_n = 5$ whenever $n$ is even and $a_n = 7$ whenever $n$ is odd. This is a number whose decimal expansion contains infinitely many $5$'s, but it also contains infinitely many non-$5$'s.

"Infinitely often" means "infinitely many times". In other words, of all of the digits in the decimal expansion of your number, infinitely many of them are $5$. It doesn't mean that all of them are $5$, or even that "most" of them are $5$ in any sense, just that there are infinitely many $5$s.