Quote (Taxidermy @ Apr 11 2010 01:01pm)
i suppose, but what i wrote is absolute convergence of the norm squared
so basically its (1-0.999...)^2 <epsilon
which is inherently a better argument than 1-0.999... < epsilon imo
but thats just me being picky probably

i see you've taken a bit of analysis

Formally, I should say |1- .999..| < epsilon to make what I wrote completely correct. My problem is that squaring it doesn't really tell you anything new. If you don't square it, what you have written is identically equal to what I have written, except that you take limits. The trouble pops up in the fact that if you don't formally construct the reals, limits aren't well defined. If you do formally construct them, then why bother taking limits at all: the answer pops out from the definitions immediately. Anyway, if you grab your copy of Rudin (I assume you took analysis with Rudin?) and look at the appendix of chapter 1, you'll see why I'm being so insistent on the above proof being the correct one.
Equivalently, we could also talk about the equivalence class of Cauchy sequences converging to 1 and use the triangle inequality on a rational approximation of .9999... I would agree that that is an equally correct "minimal" (in the sense that it does not assume any claim that could prove the result directly) answer.
This post was edited by darkfire on Apr 11 2010 12:26pm