It's too bad everyone here is too well-informed, and there's no one left to argue the other side. It's kind of fun seeing what they can come up with when backed into a corner. Just for fun, here's how I would lay out the argument:
1. [tex]0.999... = \sum_{n=1}^\infty \frac{9}{10^n}[/tex]
(by definition)
2.[tex]\sum_{n=1}^\infty \frac{9}{10^n} = \lim_{N \rightarrow \infty} \sum_{n=1}^N \frac{9}{10^n}[/tex]
(by definition)
3. [tex]\sum_{n=1}^N \frac{9}{10^n} = \frac{(9/10) - 9/(10^{N+1}) }{1-(1/10)} = 1-{\left( \frac{1}{10} \right) }^N[/tex]
(easy to show with algebra, and not many people would deny it because there aren't any infinities)
4. Let [itex]\epsilon>0[/itex]. Then there is some N with [itex](1/10)^n<\epsilon[/itex] for all n>N.
(again, pretty intuitive, doesn't involve infinites yet)
5. [tex]\lim_{N \rightarrow \infty} {\left( \frac{1}{10} \right) }^N =0[/tex]
(by (4) and the definition of a limit)
6. [tex]\lim_{N \rightarrow \infty}1- {\left( \frac{1}{10} \right) }^N =1[/tex]
(by (5) and the continuity of subtraction)
7. Therefore, by (1),(2),(3),(6), and the transitivity of equality
[tex]0.999... = 1[/tex]
Now, if they want to deny the conclusion, they have to pick a premise to deny. (1) or (2) would just be disagreeing with definitions everyone else uses, and there's no point in arguing about something like that. (3) and (7) are pretty undeniable. (4) is a little tricky, but again, there are no infinities, so I don't think it would be too hard to convince people of. (6) is actually the most technical line, but I think it jives with intuition, so I don't think it would be a problem.
That leaves us with (5). Again, debating this would just be disagreeing with a definition. The essence of the problem is that people have a preconceived notion of what a limit is, as some sort of process, but this doesn't agree with the epsilon delta definition (in fact, it doesn't really make sense at all). If they can understand this definition, then I don't see how they could both accept all the definitions used above and still deny that 0.999...=1.
