Originally posted by drnihili
Induction gets applied to all sorts of cases that aren't countable. It's pretty common actually. Most of transfinite arithmetic would evaporate without it. Every set theory book I've ever taught from or learned from does it. How do you propose to develop the theory of ordinals by only appealing to countable sets? And that just at the basic level.
As for the axiom of choice, I assume we're talking ZFC here unless otherwise specified. Sure choice is non-constructive and we can have a debate about whether it's true and such. But this isn't a thread about constructivist approaches to mathematics. So unless otherwise stated it's pretty safe to assume that the underlying set theory is something akin to ZFC.
OK, so you can use induction on any well ordered set, and by the AC, any set can be well ordered. i brought up the AC, not because i wanted to suggest that we shouldn t work in ZFC, but just because i thought that since the ordering can t be constructed, i thought it would be very difficult to actually
prove anything that way. when you write a proof by induction over the integers, you have to actually calculate the succesor of your nth term, and with an ordering that you cannot construct, i don t see how you could write a proof by induction over other sets.
however, just because i haven t seen such a proof, doesn t mean they don t exist. so i will defer to you on this.
could you say a little bit about how such an inductive proof would look?
Slow down bucko, nobody said anything about limits in the sense of analysis. I'm talking about limit ordinals here.
i have never seen limits of ordinals. the only limits i have ever seen are those found in analysis. but that doesn t mean that it can t be done, so again i defer.
If you're only using /omega as the ordinal underlying the induction (i.e. if you have no limit stage) then induction only aplies to finite n. It applies to infinitely many of them, but each of them are finite. That's my point about having proven only that no finite string of 9s after a decimal represents the same number as 1. However, unless you a finitist at heart it would be pretty odd to insist on only using /omega sequences.
OK. yes, well it was my contention that a proof by induction makes no sense to try and prove that 0.999...=1. it is not true for any finite n, and induction over the integers (\omega?) will be totally useless. i have a hard time imagining that it would get you anywhere even with bigger ordinals....
so whether or not i m a finitist, doesn t seem like it is relevant here.
my point above is that, the way the real numbers are defined, induction is not necessary. 0.999... as a real number has to be 1.
and my other point is this:
BenGooding said:
1. 0.999 is not 1 for any finite number of 9s.
2. by induction, let n be infinity. therefore 0.999... repeating is not 1.
to which StrangeDays replied:
"induction over finite n cannot be extrapolated to n=\infty"
(i paraphrase)
and i agree with StrangeDays. induction shows that 0.999 is not 1 for any finite number, and it is not valid to induct up to infinite number of 9s. i would be interested to hear how you apply transfinite induction to this argument, and i sincerely hope that you re not going to claim that it allows you to prove that 0.999... repeating does
not equal 1.
but like i said, you can make a completely rigorous argument, just using the definition of a real number, without any recourse to induction, that 0.999...=1.