1

For example, if we have a statement $f(x)$ which we have to prove whether it is true for all $x$ in the set $(0,\infty)$. Now, we first prove that $f(0)$ is true. Then, we assume $f(k)$ is true. Then we prove that $\lim_{h\rightarrow 0}f(k+h)$ true. Now, this doesn't look good to me because $\lim_{h\rightarrow 0}f(k+h)$ is exactly equal to $f(k)$ which we've assumed to be true. But, can something be done to prove a statement for all reals?

  • Have a look at transfinite induction: https://en.wikipedia.org/wiki/Transfinite_induction. – TheGeekGreek Mar 22 '17 at 10:46
  • Primitive recursive arithmetic generalizes induction fairly well. But you cannot use induction to prove a statement about reals because there is no order under which reals could be enumerated. The generalization of induction would really be to abandon induction and use other defining principles like is done for real numbers. – DanielV Mar 22 '17 at 10:51
  • It is not even true that $\lim_{h\to0}f(k+h)=f(k)$. This assumes continuity of $f$. –  Mar 22 '17 at 10:51
  • You cannot induct over reals. Assuming the axiom of choice, there is an order to the real numbers that theoretically could be induced over, but it is impossible to describe directly, and therefore is of very limited use in concrete calculations. Not assuming the axiom of choice, there is no ordering of the reals that allows induction, even theoretically. – Arthur Mar 22 '17 at 10:54
  • @arthur The reals are a subset of the hyperreals, and we induct on the hyperreals all the time. – MJD Mar 22 '17 at 12:45
  • @MJD The hyperreals cannot be constructed without the axiom of choice, usually in the form of Zorn's lemma, so there's that. I didn't know that you could induct in any meaningful way, though. – Arthur Mar 22 '17 at 12:49

0 Answers0