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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!
The order in which the terms of the sequence appear.
Since the proof you gave a link to isn't very well explained (in my opinion), I'll try to give it again here.
Theorem: For any sequence $(x_n)$ of real numbers, and an interval $[a,b]$ there is some $x\in[a,b]$ that is not a term of the sequence.
Proof: Let $a_1$ and $b_1$ be the first two terms of the sequence such that $a<a_1<b_1<b$ (if we cannot do this, then $(x_n)$ must miss infinitely many points inside $[a,b]$, so we win immediately). Now choose $a_2,b_2$ to be the next members of the sequence satisfying $a_1<a_2<b2<b1$. If we have no such $a_2,b_2$, then there must be at most one member of the sequence lying between $a_1$ and $b_1$, so we win. Otherwise, continue to build the sequence up, creating $a_1<a_2<a_3<b_3<b_2<b_1$ and so on. Eventually, we will have sequences $a_1<a_2<a_3<\dots$ and $b_1>b_2>b_3>\dots$ drawn from $(x_n)$. If, at any point, we are unable to continue, then we win immediately.
It is very important that the $a_n,b_n$ are the earliest terms in the sequence that have the desired properties. This then means that, for each $n$, $x_n$ does not belong to the interval $(a_n,b_n)$. The formal proof of this requires induction, but I'll just sketch the first few cases. If $x_1$ is one of $a_1,b_1$ then it doesn't belong to $(a_1,b_1)$. Otherwise, it must lie outside the interval $[a,b]$, so it certainly doesn't belong to $(a_1,b_1)$. Similarly, $x_2$ must lie outside $(a_2,b_2)$: either it is one of $a_1,b_1$, or it must lie outside $[a,b]$. $x_3$ must lie outside $(a_3,b_3)$, because either it is one of $a_1,b_1,a_2,b_2$ or it must lie outside $[a_1,b_1]$. Keep going like this.
Now the sequence $a_n$ is increasing and bounded above by $b$, so it tends to a limit $a_\infty$. Now suppose $a_\infty=x_n$ for some $n$. But then $a_\infty$ must belong to the interval $(a_n,b_n)$, which can't happen as we have seen. So $a_\infty$ is not a term of the sequence. $\Box$
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).
