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Let $E$ be the set of all x $\in [0,1]$ with decimal expansion containing 0s and 9s. Is E closed? Or more specifically do I consider the limit point 1 being part of set E or not?

Jia Cheng Sun
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  • I think either view is permissible. Usually people prefer a terminating form if one is available, but there is no requirement to do so. – almagest Jan 26 '20 at 10:11
  • Are you requiring only digits 0 and 9? (or only 1 and 9)? – almagest Jan 26 '20 at 10:12
  • Sorry that was confusing. I meant digits n and 9, where n is not 0 – Jia Cheng Sun Jan 26 '20 at 10:14
  • I asked that because technically 1 = 1.000000... so the decimal expansion of 1 consists of 0s, ignoring the 1 before the decimal point – Jia Cheng Sun Jan 26 '20 at 10:15
  • I don't see much point in getting too excited about this in the abstract. The basic rule is that one's math writing should be clear. So writing as tersely as possible is usually a mistake. It is better to spell things out in more detail if there is a risk of confusion. – almagest Jan 26 '20 at 10:18
  • See: https://en.wikipedia.org/wiki/0.999... – Emilio Novati Jan 26 '20 at 10:24
  • @EmilioNovati Yes I have seen it. I asked the question because I'm not sure whether the definition of the set is self-contradictory – Jia Cheng Sun Jan 26 '20 at 10:52

1 Answers1

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$1=1.000000\ldots=0.999999\ldots$ is not the only problematic value under your definition. For example

  • $\frac1{10}$ can be written $0.100000\ldots$ or $0.099999\ldots$
  • $\frac9{10}$ can be written $0.900000\ldots$ or $0.899999\ldots$

and there are many more.

So your definition needs to be clearer about what to do with these cases.

In the case of the Cantor set, similar but using digits from $\{0,2\}$ in the ternary base $3$ representation of numbers in $[0,1]$, you can say it has all numbers which can be represented with those digits, whether or not there are alternative representations which fail the condition.

If you took the same approach with your $E$ so that most conclusions about the Cantor set also applied to your $E$, then the answer to your question would be "yes" and you would have $\{1,\frac1{10}, \frac9{10}\} \subset E$. That is your choice

Henry
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  • Though note that $\frac1{11}=0.090909\ldots \in E$ anyway as this is its only decimal representation – Henry Jan 26 '20 at 13:03