TITLE: Induction on Real Numbers
QUESTION [164 upvotes]: One of my Fellows asked me whether total induction is applicable to real numbers, too ( or at least all real numbers ≥ 0) . We only used that for natural numbers so far. 
Of course you have to change some things in the inductive step, when you want to use it on real numbers.
I guess that using induction on real numbers isn't really possible, since $[r,r+\epsilon]$ with $\epsilon > 0$, $r \in \mathbb R$ is never empty.
Can you either give a good reason, why it isn't possible or provide an example where it is used?

REPLY [4 votes]: In the case of $[0,\infty[$, perhaps this looks like induction:
Let $A\subseteq [0,\infty[$ such that
$i)$ $[0,1]\subset A$
$ii)$ if $x\in A$ then $x+1\in A$.
Let $x\in[0,\infty[$. Then $x - \lfloor x\rfloor\in A$ by $i)$, and by repeating $ii)$ some  times, we get $$(x - \lfloor x\rfloor)+\underbrace{1+\cdots+1}_{\text{$\lfloor x\rfloor$ times}}=x\in A.$$ Therefore $A=[0,\infty[$.