Does the $0.\overline{9} = 1$ proof show that the gap between numbers adjacent numbers is $0$, and so are infinitesimal values equal to $0$ ? Would $10^{-\infty}=0$ ?
Edit: I feel like the smallest non-zero number would be $0.0...1$, which when subtracted from $1$ would 'appear' to be $0.999...$ , but it can't since $1 = 0.999...$
On
The real numbers do not contain infinitesimals. In systems with infinitesimals, it is not true that $0.\overline 9 = 1$, using its usual definition as series limit. Indeed, in systems with infinitesimals, $0.\overline9$ is actually undefined, as the series won't converge.
If you define $0.\overline 9$ differently, then in those systems it may or may not be equal to $1$, depending on the definition.
However note that the real numbers are complete, that is, all Cauchy sequences converge in them, and therefore there is no need to introduce infinitesimal values. Indeed, when considering convergence they are actually harmful.
You have: $0.9999...= \dfrac{9}{10}+\dfrac{9}{100}+\cdots= \dfrac{\frac{9}{10}}{1-\dfrac{1}{10}}=1$. For the other question $10^{-\infty} = \displaystyle \lim_{n \to -\infty} 10^n = 0$ ( you can prove this using definition ) .