TITLE: Cantor's First Proof that $\Bbb{R}$ is uncountable
QUESTION [5 upvotes]: I was reading this proof and I do not understand why at some point they argue that if $h$ is in the sequence then there are only finitely many terms before it. My question is by this "preceding" is it meant order of real numbers or the order in which the terms of the sequence appear?
I would very much appreciate an explanation. Thank you!

REPLY [5 votes]: If $h=x_k$, then by "the terms preceding $h$" what we mean is $x_1,x_2,\dots,x_{k-1}$, that is, terms in the sequence with index smaller than $k$. No relation with the usual order of $\mathbb R$ is required here.
The proof continues by saying that finitely many of the $a_i$ precede $h$. Naturally: Each $x_i$ is $a_i$ or $b_i$, so of the finitely many numbers $1,2,\dots,k-1$, only finitely many are such that $x_i=a_i$.  
The relation with the order of $\mathbb R$ only appears in the last paragraph of the proof: Having defined $d$ as the largest $i<k$ such that $x_i=a_i$, we have that $h\in(a_{d+1},b_{d+1})$. But then $a_{d+1}<h$, and since the $a_i$ are increasing, then $a_{d+1}$ precedes $h$ as well.
Actually, you may want to look at this beautiful (and, I think, clearer) presentation of the argument, that also reveals an unexpected connection with the golden ratio:

Mike Krebs, and Thomas Wright. On Cantor's First Uncountability Proof, Pick's Theorem, and the Irrationality of the Golden Ratio, Amer. Math. Monthly, 117 (7), (2010), 633–637. MR2681523 (2011e:11127).