TITLE: Understanding the countable ordinals up to $\epsilon_{0}$
QUESTION [37 upvotes]: in a recent MO question, link, discussing the current foundations of mathematics, the author linked a video lecture by Prof.  Voevodsky, which argues against the principle of $\epsilon_{0}$-induction used in Gentzen's proof of the consistency of PA.
In discussions arising from the question, some people commented that imagining an infinite descending chain in $\epsilon_{0}$ is "crazy".
I would like to understand better this ordinal, since I actually don't know exactly how to depict it in my mind.
I have clear in my mind the order associated with the finite ordinals. I use in my mind a notation of the following kind:
$1 = I$
$2=  II$
$3=  III$
$4=  IIII$
$\omega = (III\dots)$
$\omega+1= (III\dots)I$
$\omega +2 = (III\dots)II$
$\omega + \omega= \omega \cdot 2= (III\dots)(III\dots)$
In general I understand $\alpha + \beta$ as the juxtaposition of the two representations.
$\omega\cdot 3 = (III\dots)(III\dots)(III\dots)$
$\omega\cdot \omega = \omega^{2} = \big( (III\dots)(III\dots)(III\dots)\dots\big)$
In general I understand $\alpha \cdot \beta$, by replacing each $I$ symbol in $\beta$ with the representation of $\alpha$. So
$\omega^{3}=\omega^{2}\cdot \omega = \big( \omega^{2} \omega^{2}  \omega^{2}  \dots \big)$  
This allows me to visualize every ordinal of the form $\omega^{n}\cdot m + k$, with $n,m,k$ naturals (i.e finite ordinals). So far I have absolutely no doubt that there are no infinite descending chain in ordinals of the form $\omega^{n}\cdot m + k$.
However I start having problem with the ordinal $\omega^{\omega}= \bigsqcup_{n<\omega}\omega^{n}$. Do you have any idea on how to visualize $\omega^{\omega}$ is a way consistent with the representation used above (which i actually found here) ?
Anyway, looking at wikipedia, I still manage to visualize $\omega^{\omega}$ as the set of infinite strings of natural number, having only finitely many digits different from $0$.
Still I have no doubt that there are no infinite descending chain in  $\omega^{\omega}$.
Perhaps i might be able to understand $\omega^{\omega^{\omega}}$, namely the set of infinite strings labeled with elements of $\omega^{\omega}$, having only finitely many elements different from $0$.
Or (i guess) equivalently a $\omega\times\omega$ square labeled with naturals, where only finitely many columns are different from $0^{\omega}$, and all of these non constant-$0$ columns, contains only finitely many digits different from $0$.
However I do not know how to visualize $\epsilon_{0}$. I mean I know that the elements of $\epsilon_{0}$ can be represented by finite-branching finite trees labeled with natural numbers, but that doesn't give me a strong intuition about the fact that no infinite chain exists, so I guess its not a great picture (or at least I do not understand it properly, yet).
Questions
A) Could you suggest a way to visualize $\omega^{\omega^{\omega}}$? It should be in such a way to convince me about the fact that there are no infinite down-chain.
B) Could you suggest a way to visualize $\epsilon_{0}$, again arguing that it should be very clear that there are no infinite down-chain. 
C) Could you please state your opinion about Prof. Voevodsky, which argues against the principle of $\epsilon_{0}$-induction used in Gentzen's? 
This shouldn't be a duplicate of the previous, wider thread link, I'm only interested in this little bit of Voevodsky's talk.
Thank you in advance,
bye 
matteo

REPLY [5 votes]: In the course of writing an expository article on the consistency of arithmetic, I was led to try to explain $\epsilon_0$ in as "finitary a manner" as possible.  Here is what I came up with, inspired greatly by the account in Torkel Franzen's book Inexhaustibility: A Non-Exhaustive Account, as well as the LISP programming language.
Define a
list
to be either an empty sequence—denoted
by ( ) and referred to as the empty list—or, recursively, a
finite   nonempty   sequence   of   lists.  If we write a list by separating the
constituent elements by commas and enclosing the entire sequence with parentheses,
then   for   example,
(( ), ( ), ( ))  and  ((( ), ( )), ((( ), (( ), ( ))), ( )))  are  lists.
The number of constituent lists is called its length
(it is
zero for the empty list). If
$a$
is a nonempty list, then we
write $a[i]$ for the
$i$th constituent list of $a$,
 where $i$ ranges
from 1 to the length of $a$.
Next, recursively define a total ordering $\le$
on lists as
follows  (it is essentially  a lexicographic ordering). Let
$a$ and $b$
be lists, with lengths
$m$ and $n$
respectively. If $m\le n$ and $a[i] = b[i]$  for all $1\le i \le m$
(this condition is vacuously
satisfied if
 $m=0$), then $a\le b$. Otherwise, there exists some $i$
such that
$a[i] \ne b[i]$; let $i_0$
be the least such number, and
declare $a< b$ if $a[i_0] < b[i_0]$.
Finally, recursively define a list $a$
to be an
ordinal (or more precisely, an ordinal below $\epsilon_0$, but I will just say ordinal for short)
if all
its  constituent  lists  are  ordinals  and $a[i] \ge a[j]$ whenever $i<j$.
(In particular, the empty list is an ordinal, since the
condition is vacuously satisfied.)
As an exercise, you can verify that the ordinals ((( ))) and ((( ), ( ))) and ((( )), (( ))) are, in standard notation, the same as $\omega$, $\omega^2$, and $\omega \cdot 2$ respectively.
My personal opinion is that the usual ways of defining $\epsilon_0$ make it seem mind-bogglingly infinitary, but the above definition makes it clear that any particular ordinal is just a special type of finite list of finite lists of finite lists of … of finite lists.  Not very infinitary at all.
The proof that there is no infinite descending sequence is not particularly mind-boggling either, in this language.  Rather than type it out here, I refer you to the proof of Theorem 1 in my expository article.
I find that after understanding this proof, the hard thing to wrap my head around is how it can possibly be true that PA does not prove that there is no infinite descending sequence.  My current gut feeling is that PA seems weirdly weak, because it cannot even formalize a proof as simple as this one.