In the surreal numbers, is it fair to say $0.9$ repeating is not equal to $1$?

Question

I find the surreal numbers very interesting. I have tried my best to work through John Conway's On Numbers and Games and teach myself from some excellent online resources.

I have prepared a short video to introduce surreal numbers, but I want to double-check some of my claims and would appreciate some help.

1. Is it fair to say that $0.999$ repeating is not equal to $1$ in the surreal numbers? This is the title of my video, so I really want to make sure it's a reasonable statement.

2. Can $\{0, \frac12, \frac34, \frac78, \dots \mid 1\} = 1 - \epsilon$ be thought of as $0.999$ repeating?

3. Is the number $\{0, \frac{9}{10}, \frac{99}{100}, \frac{999}{1000}, \dots \mid 1\}$ also equal to $1 - \epsilon$?

I am hoping the the answer is "Yes" for each question. If not, please let me know and I will get on the task of majorly revising the video. Thanks!

Answer 1

The surreal numbers contain the real numbers (as well as infinite and infinitesimal numbers).

Both $0.999\dots$ and $1$ are real numbers.

In the real numbers $1 = 0.999\dots$ and so it must also be true in the surreal numbers.

As Bryan correctly points out in his answer, surreal numbers which are not real numbers do not have a decimal expansion. This would appear to undermine the idea of using $0.999\dots$ to represent a surreal number.

Answer 2

Consider the sequence $$x_n=\sum_{i=1}^n\frac{9}{10^i}$$

whose 'limit' we understand to be the thing we call $.9\bar{9}$. When we consider this sequence in the surreal numbers, it does not converge to anything.

Let $I(1)$ represent the neighborhood of all infinitesimal numbers around $1$. The sequence does not converge to $1$ because $x_n$ never enters $I(1)$. If we consider any other real number $x$ less than $1$, the sequence will eventually surpass it and also the neighborhood $I(x)$. So if it converged, it would have to converge to some infinitesimal number to the left of $1$ which is in the infinitesimal neighborhood of $1$. But again, that sequence never enters $I(1)$. So $x_n$ does not converge; that is, $.9\bar{9}$ does not represent anything meaningful in the surreal numbers.

Answer 3

I think you could say that $1 - \varepsilon$ is $0.999...$ in the sense that $\forall n \in \mathbb{N}, \left\lfloor 10^{n+1}(1-\varepsilon) \right\rfloor - 10\left\lfloor 10^n(1-\varepsilon) \right\rfloor = 9$ and $1 - \varepsilon = (+-+++...)$ is the simplest surreal satisfying this.

$1-\varepsilon = \{0;0.9;0.99;...\} | \{1\}$ because $\{0;0.9;0.99;...\}$ and $\{0;\frac{1}{2};\frac{3}{4};...\}$ are mutually cofinal.