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If 0.999.. = 1, does that mean that infinitesimals are not allowed in $(-\infty,1)$? Otherwise, we would have $0.9 \in (-\infty,1)$, $0.99 \in (-\infty,1)$, $0.999\in(-\infty,1)$, ad infinitum.

Julian Cienfuegos
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    $[0,1]$ is a closed interval. (proof?) There are many more though. – ILoveMath Dec 02 '13 at 04:16
  • "Ad infinitum"? Explain, please. – DonAntonio Dec 02 '13 at 04:16
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    Ah, los dones al ataque... – DonAntonio Dec 02 '13 at 04:16
  • atacando como siempre los .... problemas! hahaha – ILoveMath Dec 02 '13 at 04:18
  • Are you completely convinced that $0.\overline{9}=1$ and identically 1? – J. W. Perry Dec 02 '13 at 04:19
  • @Mr. Perry I am trying to figure out where I stand on this issue... I thought at first that there was no way they were the same thing, but after seeing a couple of proofs, and hearing respectable mathematicians say that the two are equal, I am convinced they are the same. I cannot get over the gut feeling that this is somehow contains a contradicition to the notion of [0,1] being closed. The best I can figure is what Abhishek says below about only terminating decimals 0.99...9 being in this interval. But that sounds wrong. – Julian Cienfuegos Dec 02 '13 at 04:31
  • @JulianCienfuegos It sounds like you are on the right track. Work the algebraic proof that $0.\overline{999}$ is identically $1$ and see that it is true. – J. W. Perry Dec 02 '13 at 04:34
  • @DonAntonio I mean "and so on forever", like continuously appending 9s to the end of that decimal forever. – Julian Cienfuegos Dec 02 '13 at 04:41

3 Answers3

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It is true that $0.999 \cdots 9$, with any finite number of nines, is in $(-\infty, 1)$. But because this interval is not a closed set, the limit of this sequence of points doesn't necessarily have to be in the set.

$0.999 \dots =1$ is not in $(-\infty, 1)$.

The set $[0,1]$ is a closed interval and a closed set. There is no sequence of points in this interval such that their limit is not in the interval.

$0.999 \dots =1$ is in $\left[\,0,1\,\right]$.

Zubin Mukerjee
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In the real number system, infinitesimals are not a thing. So no infinitesimals are allowed anywhere. There are other number systems as discussed in https://www.dpmms.cam.ac.uk/~wtg10/decimals.html but they are not standard.

comptuerbro
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0.999999.... is allowed in (−∞,1) , as whatever will be the no of 9's after decimal point it will always be less than 1 so as per definition of open interwal (−∞,1), it will always be belonging to that interwal.

Abhishek
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  • How about infinitely many 9's after the decimal point? – J. W. Perry Dec 02 '13 at 04:24
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    think open interwal as a circle or boll like structure whose boundry points are not part of the system as in this case 1 and −∞ are not the element of system . As you say if there will a alot no. of 9's after decimal , think that it would always be less than 1. when ever If it is 1 it will not be part of the system. – Abhishek Dec 02 '13 at 04:33