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?

up vote 4 down vote accepted

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.

  • So, in other words, there is some infinite "string" of $5$'s in each number in this set of real numbers? – Thy Art is Math Sep 22 at 5:09
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    @ThyArtisMath Yes, there is some infinite string 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. – Brahadeesh Sep 22 at 5:13
  • Got it! Thank you for the example -- that makes sense to me now! – Thy Art is Math Sep 22 at 5:15
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    @ThyArtisMath Glad to be of help. :) – Brahadeesh Sep 22 at 5:18

"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.

  • Ah, that makes sense. Thank you for the clarification. Would you happen to have any suggestions for a way in which I can explicitly write such a number mathematically? Or do you think that writing that out would be unhelpful in this case? – Thy Art is Math Sep 22 at 5:05
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    I doubt such a representation would be helpful. That would be essentially writing your set as the image of some function, but that is not likely to be helpful for showing it is Borel since an image of a Borel set under a Borel function need not be Borel – Eric Wofsey Sep 22 at 5:08
  • Ah ok, I did not know that. Thanks for your feedback! – Thy Art is Math Sep 22 at 5:13

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