How can $0.999\dots$ not equal $1$?

First, by definition I assume that $0.999...$ actually is defined as: $$\text{lim}_{n\rightarrow\infty}\sum_{i=1}^n 9/10^i$$

Now by geometric series we already know that this equals one. But nonetheless here is an explicit proof. The statement is: $$\forall\epsilon_+\exists\delta\forall n (n>\delta\rightarrow|\sum_{i=1}^n 9/10^i-1|<\epsilon)$$

Which is equivalent to:

$$\forall\epsilon_+\exists\delta\forall n (n>\delta\rightarrow|\sum_{i=0}^n 9/10^i-10|<\epsilon)$$

Let $\epsilon>0$ be a real number. Now note that $\sum_{i=0}^{n} 9/10^i=10-1/10^n$. Choose $\delta=\text{max}(1,\text{ceil(log}(1/\epsilon)))$

As such: $|10-10-1/10^n|=1/10^n<1/10^{\text{ceil(log}(1/\epsilon))}\leq 1/10^{\text{log}(1/\epsilon)}=\epsilon$

Which works out nice...

However, I have learned that $0.999...=1$ doesn't hold in all number systems such as hyperreals and surreals and what not. I am not even sure about rational numbers (although the proof looks like it would work for rationals with small tweaks)... I believe that the statement can't even be formulated in the first order theory of real closed fields as the $n$ is quantified over naturals, so there are number systems which can't even express the fact.

What in the proof goes wrong in the non-standard number systems and what are the primary features of the systems that cause this? Also, a proof of the fact NOT holding in said systems is welcome!

Solution 1:

$0.\bar{9}= 1$ does hold the hyperreals.

The fact you're referring to is something different, and unfortunately that difference is usually not made clear which leads to confusion like you had. What people are trying to say is that in the hyperreals, you can have a terminating decimal that nonetheless has infinitely many $9$'s.

More precisely, you still have decimal notation for the hyperreals, but the places are indexed by hyperintegers rather than ordinary integers. And if you take an infinite hyperinteger $H$, then

$$ 1 - 10^{-H} = \sum_{n=1}^H 9 \cdot 10^{-n} = 0.\underbrace{999\ldots999}_{H\text{ nines}} \neq 1$$

However, $0.\bar{9}$ still refers to the nonterminating decimal that has a $9$ in every fractional place, and $0.\bar{9} = 1$.