Fundamental sequences for the Veblen hierarchy of ordinals

[jacquesb]

I will first summarize the construction of ordinal numbers and introduce the definition of the binary Veblen function and of the notion of fundamental sequence.

Ordinal numbers start with natural numbers 0, 1, 2, 3, ... which are followed by $\omega$ which represents the "simple" infinity. Then it goes further with $\omega+1, \omega+2, \omega+3, ... \omega+\omega = \omega \cdot 2, \omega \cdot 2 + 1, ... \omega \cdot 3, ... \omega \cdot \omega = \omega^{\omega}, ..., \omega^{\omega^{\omega}}, ..., \epsilon_0, ...$.

$\epsilon_0$ is the least fixed point of the function $\alpha \rightarrow \omega^{\alpha}$. It is the limit of the sequence starting with 0 and repeating application of this function : $0, \omega^0, \omega^{\omega^0}, ...$

The next fixed point, written $\epsilon_1$, is the limit of the sequence starting with the next ordinal $\epsilon_0+1$ : $\epsilon_0+1, \omega^{\epsilon_0+1}, \omega^{\omega^{\epsilon_0+1}}, ...$

The following fixed points $\epsilon_2, \epsilon_3, ...$ can be defined the same way.

We can say that the function $\alpha \rightarrow \epsilon_\alpha$ enumerates the fixed points of the function $\alpha \rightarrow \omega^\alpha$.

Then $\epsilon_\omega$ can be defined as the limit of $\epsilon_0, \epsilon_1, \epsilon_2, \epsilon_3, ...$.

And we can go on with $\epsilon_{\omega+1}, ..., \epsilon_{\epsilon_0}, ... \epsilon_{\epsilon_{\epsilon_0}}, ..., \zeta_0, ...$.

$\zeta_0$ is the least fixed point of the function $\alpha \rightarrow \epsilon_\alpha$ .

In a similar way, the function $\alpha \rightarrow \zeta_\alpha$ enumerates the fixed points of the function $\alpha \rightarrow \epsilon_\alpha$.

We can go on using successive greek letters, but it is simpler to number these functions, defining :

. $\varphi_0(\alpha) = \omega^\alpha$
. $\varphi_1(\alpha) = \epsilon_\alpha$
. $\varphi_2(\alpha) = \zeta_\alpha$

and so on.

The binary Veblen function $\varphi_{\alpha}\beta$, sometimes written $\varphi(\alpha,\beta)$ or $\varphi \ \alpha \ \beta$, is defined by (see for example http://www.cs.man.ac.uk/~hsimmons/TEMP/OrdNotes.pdf):

. $\varphi_0(\alpha) = \omega^{\alpha}$
. $\varphi_{\alpha+1} =$ enumeration of fixed points of $\varphi_{\alpha}$
. $\varphi_{\lambda} =$ enumeration of common fixed points of $\varphi_{\alpha}$ for all $\alpha < \lambda$ (where $\lambda$ is a limit ordinal)

See also https://sites.google.com/site/largenumbers/home/4-2/fgh_gamma0 and http://quibb.blogspot.fr/2012/03/infinity-larger-countable-ordinals.html for more information.

A sequence of ordinals $\alpha_0, \alpha_1, \alpha_2, ...$ is said to be a fundamental sequence of the limit ordinal $\lambda$ if $\lambda$ is the limit of $\alpha_0, \alpha_1, \alpha_2, ...$.

The fundamental sequences of $\phi_{\beta}\gamma$ are defined by (see for example https://en.wikipedia.org/wiki/Veblen_function):

. For any $\beta$, if $\gamma$ is a limit with $\gamma <\varphi _{{\beta }}(\gamma )$ then let $\varphi _{{\beta }}(\gamma )[n]=\varphi _{{\beta }}(\gamma [n])$.
. No such sequence can be provided for $\varphi _0(0) = \omega^0 = 1$ because it does not have cofinality $\omega$.
. For $\varphi _0(\gamma +1)=\omega ^{{\gamma +1}}=\omega ^{\gamma }\cdot \omega \,,$ we choose $\varphi _0(\gamma +1)[n]=\varphi _0(\gamma )\cdot n=\omega ^{{\gamma }}\cdot n\,$.
. For $\varphi _{{\beta +1}}(0)$, we use $\varphi _{\beta +1}(0)[0]=0$ and $\varphi _{{\beta +1}}(0)[n+1]=\varphi _{{\beta }}(\varphi _{{\beta +1}}(0)[n])\,,$ i.e. 0, $\varphi _{{\beta }}(0)$, $\varphi _{{\beta }}(\varphi _{{\beta }}(0))$, etc.
. For $\varphi _{{\beta +1}}(\gamma +1)$, we use $\varphi _{\beta +1}(\gamma +1)[0]=\varphi _{\beta +1}(\gamma )+1$ and $\varphi _{{\beta +1}}(\gamma +1)[n+1]=\varphi _{{\beta }}(\varphi _{{\beta +1}}(\gamma +1)[n])\,$.

Now suppose that $\beta$ is a limit:
. If $\beta <\varphi _{{\beta }}(0)$, then let $\varphi _{{\beta }}(0)[n]=\varphi_{{\beta [n]}}(0)\,$.
. For $\varphi _{{\beta }}(\gamma +1)$, use $\varphi _{{\beta }}(\gamma +1)[n]=\varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma )+1)\,$.

It seems rather clear to me that these sequence are effectively fundamental sequences of $\varphi_\beta(\gamma)$, except for the last one. I don't understand why $\varphi _{{\beta [n]}}$ is applied only once to $\varphi _{{\beta }}(\gamma )+1$, it seems to me that it should be applied repeatedly to get a fixed point.

Could someone explain this to me, or indicate me where I could find a proof that these sequences are fundamental sequences of $\varphi_\beta(\gamma)$ ?

 

[SSequence]

Now suppose that $\beta$ is a limit:
For $\varphi _{{\beta }}(\gamma +1)$, use $\varphi _{{\beta }}(\gamma +1)[n]=\varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma )+1)\,$.

It seems rather clear to me that these sequence are effectively fundamental sequences of $\varphi_\beta(\gamma)$, except for the last one. I don't understand why $\varphi _{{\beta [n]}}$ is applied only once to $\varphi _{{\beta }}(\gamma )+1$, it seems to me that it should be applied repeatedly to get a fixed point.

Could someone explain this to me, or indicate me where I could find a proof that these sequences are fundamental sequences of $\varphi_\beta(\gamma)$ ?

It seems that your question is about the equation quoted above (I won't comment about the other equations as each will have to be looked carefully). I don't think this is immediately clear at all. However, it does seem reasonably plausible on a first look.

The thing is that when you are dealing with $\beta$ as a limit, it might help to look at things more slowly. Think of $\varphi _{{\beta }}(\gamma)$. As it was mentioned because it is a common fixed point for all functions $\varphi _{{\alpha }}$ (where $\alpha<\beta$), we have by definition:
$\varphi _{{\alpha }}(\varphi _{{\beta }}(\gamma)) = \varphi _{{\beta }}(\gamma)$ for all $\alpha<\beta$

So naturally we have:
$sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)) \,|\, n \in \omega\}=\varphi _{{\beta }}(\gamma )$

Plausibly, I think this is why in the parameter of the second equation there is $\varphi _{{\beta }}(\gamma )+1$ (seems to be a natural thing to try) instead of $\varphi _{{\beta }}(\gamma)$. Of course this shouldn't be taken as a rigorous argument by itself.

================

Generally speaking, I think you have to be careful on limits. The reason is that the function $\varphi _{\beta }$ (when $\beta$ is a limit) isn't defined directly by point-wise limits. That is, in general the following equation is not true (here $\beta$ is a limit) for all values of $\gamma$:
$sup\{ \varphi _{{\beta [n]}}(\gamma) \,|\, n \in \omega\}=\varphi _{{\beta }}(\gamma)$

Let $\beta$ be a limit. Define a function obtained by point-wise limits as:
$f_{\beta}(x)=sup\{ \varphi _{{\alpha}}(x) \,|\, \alpha<\beta\}$
Basically if you remove the repetitions in the above function $f_{\beta}$ (generally speaking at least, this function won't be normal because it won't be strictly increasing), I think you will get the function $\varphi _{{\beta }}(x)$ (though I have never tried that) ... where $\beta$ is a limit.
One might also observe that if this alternative definition is taken (and proven to be equivalent) then the defining equation (for limit $\beta$) $\varphi _{{\beta }}(\gamma +1)[n]=\varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma )+1)\,$ also makes a lot of sense (shouldn't be difficult to see).

$\epsilon_0$ is the least fixed point of the function $\alpha \rightarrow \omega^{\alpha}$. It is the limit of the sequence starting with 0 and repeating application of this function : $0, \omega^0, \omega^{\omega^0}, ...$

The next fixed point, written $\epsilon_1$, is the limit of the sequence starting with the next ordinal $\epsilon_0+1$ : $\epsilon_0+1, \omega^{\epsilon_0+1}, \omega^{\omega^{\epsilon_0+1}}, ...$

The sequence for $\epsilon_1$ above should be correct I believe (as long as one doesn't mix the implied precedence in the exponentiation). I think when you evaluate the limit of this sequence, it will turn out to be a tower $\epsilon_0$'s (I remember doing a calculation quite sometime ago where calculation of first three or four terms seemed to suggest that ... but I didn't confirm or double check it enough though ... can be proved by ordinary induction ofc). That is:
$\epsilon_1 = sup\{ \epsilon_0, \epsilon_0^{\epsilon_0}, \epsilon_0^{\epsilon_0^{\epsilon_0}} ... \}$

One thing here to note is that you are informally equating fixed points and limit/sup of nested compositions (of a normal function). That is true of course, but if one wants to be more rigorous that also has to be proved ... probably shown in reference(1) below.

Could someone explain this to me, or indicate me where I could find a proof that these sequences are fundamental sequences of $\varphi_\beta(\gamma)$ ?

Three references that I know of that might help in a more rigorous understanding:
1-- the original paper of Oswald Veblen
2-- Lectures on Proof Theory (Second Chapter) --- W.W. Tait
3-- Proof Theory (Fifth Chapter) -- Kurt Schutte

The first two of these should be freely available. I haven't read any of these so I can't confirm how easy or difficult it is to read/understand these.

There are probably number of other resources, but they seem to be concerned with much higher elements usually. Of course this is as far as I know, so I still might have missed some good references.

P.S.
@Deedlit is an expert on these matters here and might provide with a proof (or reference of one) where I have just used handwaving.

Edit:

Ordinal numbers start with natural numbers 0, 1, 2, 3, ... which are followed by $\omega$ which represents the "simple" infinity. Then it goes further with $\omega+1, \omega+2, \omega+3, ... \omega+\omega = \omega \cdot 2, \omega \cdot 2 + 1, ... \omega \cdot 3, ... \omega \cdot \omega = \omega^{\omega}, ..., \omega^{\omega^{\omega}}, ..., \epsilon_0, ...$.

The equation $\omega \cdot \omega = \omega^{\omega}$ is not correct. The correct equations are:
$\omega \cdot \omega = \omega^2$
$\omega^{\omega}=sup\{ \omega^n \,|\, n \in \omega\}$

 

[SSequence]

The thing is that when you are dealing with $\beta$ as a limit, it might help to look at things more slowly. Think of $\varphi _{{\beta }}(\gamma)$. As it was mentioned because it is a common fixed point for all functions $\varphi _{{\alpha }}$ (where $\alpha<\beta$), we have by definition:
$\varphi _{{\alpha }}(\varphi _{{\beta }}(\gamma)) = \varphi _{{\beta }}(\gamma)$ for all $\alpha<\beta$

So naturally we have:
$sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)) \,|\, n \in \omega\}=\varphi _{{\beta }}(\gamma )$

Plausibly, I think this is why in the parameter of the second equation there is $\varphi _{{\beta }}(\gamma )+1$ (seems to be a natural thing to try) instead of $\varphi _{{\beta }}(\gamma)$. Of course this shouldn't be taken as a rigorous argument by itself.
================

Just for the sake of some further clarity on this point (in case one doesn't want to use directly the discussion in the paragraph that follows) ... if we want to verify the following equation (for limit $\beta$):
$\varphi _{{\beta }}(\gamma +1)[n]=\varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma )+1)\,$

then we just want to show the following two points:
(i) $\varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma )+1)$ is a strictly increasing sequence of values (for increasing n).
(ii) $sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, n \in \omega\}=\varphi _{{\beta }}(\gamma+1 )$

To show (ii) ---- since $\varphi _{\beta }$ enumerates all of the common fixed points (in strictly increasing order) of all $\varphi _{\alpha }$'s (where $\alpha<\beta$), we can do this by proving the following two facts:
(1) $sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, n \in \omega\}$ is a common fixed point for all $\varphi _{\alpha }$'s (where $\alpha<\beta$).
(2) There is no value between $\varphi _{{\beta }}(\gamma)$ and $sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, n \in \omega\}$ that is a common fixed point for all $\varphi _{\alpha }$'s (where $\alpha<\beta$).

 

[jacquesb]

Just for the sake of some further clarity on this point (in case one doesn't want to use directly the discussion in the paragraph that follows) ... if we want to verify the following equation (for limit $\beta$):
$\varphi _{{\beta }}(\gamma +1)[n]=\varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma )+1)\,$

then we just want to show the following two points:
(i) $\varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma )+1)$ is a strictly increasing sequence of values (for increasing n).
(ii) $sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, n \in \omega\}=\varphi _{{\beta }}(\gamma+1 )$

To show (ii) ---- since $\varphi _{\beta }$ enumerates all of the common fixed points (in strictly increasing order) of all $\varphi _{\alpha }$'s (where $\alpha<\beta$), we can do this by proving the following two facts:
(1) $sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, n \in \omega\}$ is a common fixed point for all $\varphi _{\alpha }$'s (where $\alpha<\beta$).
(2) There is no value between $\varphi _{{\beta }}(\gamma)$ and $sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, n \in \omega\}$ that is a common fixed point for all $\varphi _{\alpha }$'s (where $\alpha<\beta$).

Thanks to SSequence for these explanations.

(i) seems intuitively plausible to me, and also (2), the less evident is (1), but I have perhaps an idea of a possible proof.

We have to prove that $sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, n \in \omega\}$ is a common fixed point for all $\varphi _{\alpha }$'s (where $\alpha<\beta$).

I guess this is the same as proving that it is a common fixed point for all $\varphi_{\beta[p]}$ with $p \in \omega$ (is it true ?)

Then we have to prove that for any $p \in \omega$, we have :

$\varphi_{\beta[p]} (sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, n \in \omega\}) = sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, n \in \omega\}$

We have for a given p :

$\varphi_{\beta[p]} (sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, n \in \omega\})$

$= \varphi_{\beta[p]} ( \varphi_{\beta[0]}(\varphi _{{\beta }}(\gamma)+1) \ \cup \ \varphi_{\beta[1]}(\varphi _{{\beta }}(\gamma)+1) \ \cup \ ... \ \cup \ \varphi_{\beta[p]}(\varphi _{{\beta }}(\gamma)+1) \ \cup \ \varphi_{\beta[p+1]}(\varphi _{{\beta }}(\gamma)+1) \ \cup \ ... )$

$= \varphi_{\beta[p]} ( \varphi_{\beta[0]}(\varphi _{{\beta }}(\gamma)+1)) \ \cup \ \varphi_{\beta[p]} (\varphi_{\beta[1]}(\varphi _{{\beta }}(\gamma)+1)) \ \cup \ ... \ \cup \ \varphi_{\beta[p]} ( \varphi_{\beta[p]}(\varphi _{{\beta }}(\gamma)+1)) \ \cup \ \varphi_{\beta[p]} ( \varphi_{\beta[p+1]}(\varphi _{{\beta }}(\gamma)+1)) \ \cup \ ...$

But since we have $\varphi_\alpha(\varphi_\beta(\gamma)) = \varphi_\beta(\gamma)$ if $\alpha < \beta$, and $\beta[p] < \beta[p+1]$, then $\varphi_{\beta[p]} ( \varphi_{\beta[p+1]}(\varphi _{{\beta }}(\gamma)+1)) = ( \varphi_{\beta[p+1]}(\varphi _{{\beta }}(\gamma)+1)$, idem for the following terms.

So the previous expression is equivalent to :

$\varphi_{\beta[p]} ( \varphi_{\beta[0]}(\varphi _{{\beta }}(\gamma)+1)) \ \cup \ \varphi_{\beta[p]} (\varphi_{\beta[1]}(\varphi _{{\beta }}(\gamma)+1)) \ \cup \ ... \ \cup \ \varphi_{\beta[p]} ( \varphi_{\beta[p]}(\varphi _{{\beta }}(\gamma)+1)) \ \cup \ \varphi_{\beta[p+1]}(\varphi _{{\beta }}(\gamma)+1) \ \cup \ ...$

We have to prove it is equal to :

$sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, n \in \omega\} = \varphi _{{\beta [0]}}(\varphi _{{\beta }}(\gamma)+1) \ \cup \ \varphi _{{\beta [1]}}(\varphi _{{\beta }}(\gamma)+1) \ \cup \ ... \ \cup \ \varphi _{{\beta [p]}}(\varphi _{{\beta }}(\gamma)+1) \ \cup \ \varphi _{{\beta [p+1]}}(\varphi _{{\beta }}(\gamma)+1) \ \cup _ ...$

These two expressions differs only by the n+1 first terms which are the smallest, the following terms ( $\varphi _{{\beta [p+1]}}(\varphi _{{\beta }}(\gamma)+1)$ and after) are equal. It seems plausible to me that the difference of the first terms does not matter, and that the equality of the following terms which are greater is sufficient for the equality of the complete expressions. What do you think about it ?

 

[SSequence]

Thanks to SSequence for these explanations.

(i) seems intuitively plausible to me...

(i) can be proved quite directly. Re-stating it (for convenience):

(i) $\varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma )+1)$ is a strictly increasing sequence of values (for increasing n).

This follows from the statement described next. When $\beta$ is a limit:
(A)-- $\varphi _{{\alpha_1}}(\varphi _{{\beta }}(\gamma )+1)>\varphi _{{\alpha_2}}(\varphi _{{\beta }}(\gamma )+1)$ for any arbitrary $\alpha_1 \, , \, \alpha_2$ where $\beta>\alpha_1>\alpha_2$

More specifically if we can prove for all $X<\beta$ that:
$\varphi _{{\alpha_1}}(\varphi _{{\beta }}(\gamma )+1)>\varphi _{{\alpha_2}}(\varphi _{{\beta }}(\gamma )+1)$ for any arbitrary $\alpha_1 \, , \, \alpha_2$ where $X \ge \alpha_1>\alpha_2$
then we are done.

We want to prove this by (transfinite) induction. As the base case, what we want to show is:
$\varphi _{1}(\varphi _{{\beta }}(\gamma )+1)>\varphi _{0}(\varphi _{{\beta }}(\gamma )+1)=\omega.\varphi _{{\beta }}(\gamma )+\omega$
We also know that (by reasoning mentioned in post#2):
$\varphi _{0}(\varphi _{{\beta }}(\gamma ))=\varphi _{1}(\varphi _{{\beta }}(\gamma ))=\varphi _{{\beta }}(\gamma )$

So we just want to show that there is no value lying between $\varphi _{{\beta }}(\gamma)+1$ (inclusive) and $\varphi _{0}(\varphi _{{\beta }}(\gamma )+1)$ (inclusive) that is a fixed point of $\varphi _{0}$. I think it shouldn't be difficult at all (but probably a bit tedious to write here in detail).

For successor case what we want to prove is:
$\varphi _{{\alpha_1}}(\varphi _{{\beta }}(\gamma )+1)>\varphi _{{\alpha_2}}(\varphi _{{\beta }}(\gamma )+1)$ for any arbitrary $\alpha_1 \, , \, \alpha_2$ where $\alpha+1 \ge \alpha_1>\alpha_2$

Our inductive assumption is:
$\varphi _{{\alpha_1}}(\varphi _{{\beta }}(\gamma )+1)>\varphi _{{\alpha_2}}(\varphi _{{\beta }}(\gamma )+1)$ for any arbitrary $\alpha_1 \, , \, \alpha_2$ where $\alpha \ge \alpha_1>\alpha_2$

It should suffice to prove that for any $\alpha<\beta$ (because proving for all combinations follows rather directly from that):
$\varphi _{{\alpha+1}}(\varphi _{{\beta }}(\gamma)+1)>\varphi _{{\alpha}}(\varphi _{{\beta }}(\gamma )+1)$
This should be nearly identical (to base case) because we can observe (somewhat trivially) that for all $x\, , \, \alpha_1 \, , \, \alpha_2$:
$\varphi _{\alpha_2}(x) \ge \varphi _{\alpha_1}(x)$ whenever $\alpha_2 \ge \alpha_1$
More specifically, for our purposes I think we can use:
$\varphi _{\alpha}(x) \ge \varphi _{0}(x)$

So a similar reasoning that we applied in base case can be carried over (with minor change).

For limit case let $\lambda$ denote some limit value less than $\beta$. What we want to prove is:
$\varphi _{{\alpha_1}}(\varphi _{{\beta }}(\gamma )+1)>\varphi _{{\alpha_2}}(\varphi _{{\beta }}(\gamma )+1)$ for any arbitrary $\alpha_1 \, , \, \alpha_2$ where $\lambda \ge \alpha_1>\alpha_2$

It should suffice to prove that (proving for all combinations should follow directly after this):
$\varphi _{{\lambda}}(\varphi _{{\beta }}(\gamma )+1)>\varphi _{{\alpha}}(\varphi _{{\beta }}(\gamma )+1)$ for any arbitrary $\alpha<\lambda$

We can show this by contradiction by assuming:
(B) -- $\varphi _{{\lambda}}(\varphi _{{\beta }}(\gamma )+1) \le \varphi _{{\alpha}}(\varphi _{{\beta }}(\gamma )+1)$ for "some" $\alpha<\lambda$

But we have (proved before):
(C) -- $\varphi _{{\alpha+1}}(\varphi _{{\beta }}(\gamma)+1)>\varphi _{{\alpha}}(\varphi _{{\beta }}(\gamma )+1)$
AND (once again by reasoning mentioned in beginning of post#2):
$\varphi _{\lambda}(\varphi _{{\beta }}(\gamma ))=\varphi _{\alpha}(\varphi _{{\beta }}(\gamma ))=\varphi _{\alpha+1}(\varphi _{{\beta }}(\gamma ))=\varphi _{{\beta }}(\gamma )$

This leads to a contradiction because, after the value $\varphi _{{\beta }}(\gamma )$, the very next fixed point of the function $\varphi _{\alpha}$ must occur after $\varphi _{{\alpha}}(\varphi _{{\beta }}(\gamma )+1)$ (otherwise equation(C) wouldn't make any sense).
However, this contradicts with equation(B) (shouldn't be too difficult to see why). That's because every value in range of $\varphi _{\lambda}$ must be a fixed point of ALL $\varphi _{X}$'s (with $X<\lambda$). And yet the value $\varphi _{{\lambda}}(\varphi _{{\beta }}(\gamma )+1)$ isn't a fixed point of $\varphi _{\alpha}$. Hence our conclusion follows.

We have to prove that $sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, n \in \omega\}$ is a common fixed point for all $\varphi _{\alpha }$'s (where $\alpha<\beta$).

I guess this is the same as proving that it is a common fixed point for all $\varphi_{\beta[p]}$ with $p \in \omega$ (is it true ?)

Yes, it should be true I think.

I haven't read in detail what you have written, but it seems you might be making (1) more complicated than it is. Re-stating it for convenience:

(1) $sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, n \in \omega\}$ is a common fixed point for all $\varphi _{\alpha }$'s (where $\alpha<\beta$).

We can observe that:
$sup\{ \varphi _{{\beta [n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, n \in \omega\}=sup\{ \varphi _{{\alpha}}(\varphi _{{\beta }}(\gamma)+1) \,|\, \alpha<\beta \}$
Note that this equality uses statement(A) above.

Now fix any arbitrary value(limit or not doesn't matter) $\lambda<\beta$. We want to prove that $sup\{ \varphi _{{\alpha}}(\varphi _{{\beta }}(\gamma)+1) \,|\, \alpha<\beta \}$ is a fixed point of $\varphi _{{\lambda}}$. Alternatively:
$sup\{ \varphi _{{\alpha}}(\varphi _{{\beta }}(\gamma)+1) \,|\, \alpha<\beta \} \in range(\varphi _{{\lambda+1}})$

Further we can observe that:
(D)-- $range(\varphi _{\alpha_1}) \supseteq range(\varphi _{\alpha_2})$ whenever $\alpha_1<\alpha_2$

Now indeed there is some $N \in \omega$ such that:
$\beta [n] \ge \lambda+1$ for all $n \ge N$ (where $n \in \omega$)
Note that because $\lambda<\beta$ and $\beta$ is a limit, we must have $\lambda+1<\beta$.

Now we can write:
$sup\{ \varphi _{{\alpha}}(\varphi _{{\beta }}(\gamma)+1) \,|\, \alpha<\beta \}=sup\{ \varphi _{{\alpha}}(\varphi _{{\beta }}(\gamma)+1) \,|\, \beta[N] \le \alpha<\beta \}=sup\{ \varphi _{{\beta[n]}}(\varphi _{{\beta }}(\gamma)+1) \,|\, N \le n < \omega \}$
Observe that the equalities use statement (A).

Now if you observe the sequence of terms:
$\varphi_{\beta[N]}(\varphi _{\beta }(\gamma)+1), \varphi_{\beta[N+1]}(\varphi _{\beta }(\gamma)+1),\varphi_{\beta[N+2]}(\varphi _{\beta }(\gamma)+1), \varphi_{\beta[N+3]}(\varphi _{\beta }(\gamma)+1),...$
This is a strictly increasing sequence of values and furthermore every term in this sequence belongs to $range(\varphi _{{\lambda+1}})$(due to statement-D above). Now let us suppose that:
$\varphi _{{\lambda+1}}(a_0)=\varphi_{\beta[N]}(\varphi _{\beta }(\gamma)+1)$
$\varphi _{{\lambda+1}}(a_1)=\varphi_{\beta[N+1]}(\varphi _{\beta }(\gamma)+1)$
$\varphi _{{\lambda+1}}(a_2)=\varphi_{\beta[N+2]}(\varphi _{\beta }(\gamma)+1)$
and so on...

Now the sequence of values $a_0,a_1,a_2,a_3...$ must form a fundamental sequence (due to $\varphi _{{\lambda+1}}$ being a strictly increasing function). Hence the limit/sup of:
$\varphi_{\beta[N]}(\varphi _{\beta }(\gamma)+1), \varphi_{\beta[N+1]}(\varphi _{\beta }(\gamma)+1),\varphi_{\beta[N+2]}(\varphi _{\beta }(\gamma)+1), \varphi_{\beta[N+3]}(\varphi _{\beta }(\gamma)+1),...$
belongs to $range(\varphi _{{\lambda+1}})$ (because $\varphi _{{\lambda+1}}$ is a normal, and hence continuous, function).

 

[SSequence]

placeholder for few comments:
--- firstly the convention for $\varphi_{0}$ seems to differ a bit (I think) since sometimes by convention people seem to denote $\varphi_{0}$ for $\epsilon_x$ instead of $\omega^x$.

--- secondly there might be some errors partly because this was my first time writing these arguments and partly because mathtyping renders somewhat slowly (especially in this quantity) on my pc. Anyway, I will try to check for any major errors (before the edit time runs out).

--- I think I have probably made a bit of a mess of induction proof for (i) (the earlier part of post#5) by choosing a more complex statement than needed. But anyway...

 

[SSequence]

(i) can be proved quite directly. Re-stating it (for convenience):

(A)-- $\varphi _{{\alpha_1}}(\varphi _{{\beta }}(\gamma )+1)>\varphi _{{\alpha_2}}(\varphi _{{\beta }}(\gamma )+1)$ for any arbitrary $\alpha_1 \, , \, \alpha_2$ where $\beta>\alpha_1>\alpha_2$

More specifically if we can prove for all $X<\beta$ that:
$\varphi _{{\alpha_1}}(\varphi _{{\beta }}(\gamma )+1)>\varphi _{{\alpha_2}}(\varphi _{{\beta }}(\gamma )+1)$ for any arbitrary $\alpha_1 \, , \, \alpha_2$ where $X \ge \alpha_1>\alpha_2$
then we are done.

We want to prove this by (transfinite) induction. As the base case, what we want to show is:
$\varphi _{1}(\varphi _{{\beta }}(\gamma )+1)>\varphi _{0}(\varphi _{{\beta }}(\gamma )+1)=\omega.\varphi _{{\beta }}(\gamma )+\omega$

Brief comment is needed here for the last equation. I took (probably because it was the easiest continuous function in my mind):
$\varphi _{0}(x)=\omega \cdot x$

But if we want to preserve the usual conventions we should take $\omega+\omega \cdot x$. The standard is to use $\omega^x$ or $\epsilon_x$ (I don't know which one is more preferred among these two). One would need to look a little more carefully to ascertain whether this kind of choice (between $\omega+\omega \cdot x$ and $\omega^x$) affects the first fixed point of $\varphi _{x}(0)$.