What is meant by "infinitely often" in this problem on Borel sets? 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 $\mathcal{A}_5$ denote the set of real numbers that have a decimal expansion with the digit $5$ appearing infinitely often, i.e. every $a\in\mathcal{A}_5$ can be written as $$a=\ldots+ 5\cdot10^n+5\cdot10^{n-1}+\ldots+5\cdot 10 +5+5\cdot10^{-1}+\ldots 5\cdot10^{-m-1}+5\cdot10^{-m}+\ldots$$

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