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I read the weakly Mahlo ordinal is weakly inaccessible , hyper-weakly inaccessible, hyper-hyper-weakly inaccessible, (1@α)-weakly inaccessible, and so on as far as you diagonalize.

But is it the first cardinal with this property?

More concretely if you "define" $$a_0(x)=x+1$$ $$a_\alpha(x)=\text{the $x^{th}$ ordinal in } \{y\mid\gamma<y,\beta<\alpha,a_\beta(\gamma)<y,y\text{ is regular if $x$ is a succesor}\}$$ $$\text{if }\operatorname{cf}(\alpha)<N$$ $$a_\alpha(x)=a_{\alpha[x]}(x)\text{ if }\operatorname{cf}(\alpha)=N$$ $$N=\min\{y\mid \gamma<y,\operatorname{cf}(\beta)\le N,a_\beta(\gamma)<y\}$$

$N$ is the smallest ordinal unreachable using inaccessibility. But is this the Mahlo ordinal?

What if you add the restriction $N$ is regular?

If it is not the first Mahlo ordinal could you use it (with regularity) in a Mahlo OCF?

Analysis of $a$

$a_1(x)=\omega_x$
$a_2(x)=I_x$
$a_{2+\alpha} : \alpha\text{-inaccible}$
$a_N(2+x)=a_{2+x}(x)=\text{ the ${2+x}^{th}$ $x$-inaccessible}$
$a_{N+1} : (1,0)\text{-inaccible}$
$a_{N+N+1} : (2,0)\text{-inaccible}$
$a_{N\cdot\alpha+1} : (\alpha,0)\text{-inaccible}$
$a_{N^2+1} : (1,0,0)\text{-inaccible}$
$a_{N^2\cdot\alpha+1} : (\alpha,0,0)\text{-inaccible}$
$a_{N^3+1} : (1,0,0,0)\text{-inaccible}$
$a_{N^\alpha +1} : (1@\alpha)\text{-inaccible}$

It's clear $N$ is used as a diagonaliser, so $N$ is self must be larger than any diagonalisation of $I$, like $M$

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  • $\begingroup$ You define $N$ in terms of itself. $\endgroup$
    – Asaf Karagila
    Commented May 25, 2018 at 16:54
  • $\begingroup$ the $\operatorname{cf}(\beta)\le N$ part is only to ensure $a_\beta$ is defined. But yes the defenition seems to be circular which is why I wrote "define", is it possible $N$ is still defined by this even is it seems circular (like $2x=x+1$ )? $\endgroup$
    – fejfo
    Commented May 25, 2018 at 17:01
  • $\begingroup$ JDH explains why the answer is a definitive "no" in his answer to this question: math.stackexchange.com/questions/24505/… $\endgroup$
    – Deedlit
    Commented Jun 5, 2018 at 8:44

1 Answer 1

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I can't say I completely follow your definition, but you might want to look at the following paper:

Carmody, Erin, Killing them softly: degrees of inaccessible and Mahlo cardinals, Math. Log. Quart. 63: 256--264, Article ID https://doi.org/10.1002/malq.201500071, (2017).

She develops a useful notation system for the degrees of inaccessibility (like hyper-inaccessible etc.) and, among other things, investigates the relationship of these to Mahlo cardinals (see especially Section 3). In particular, she shows that it is consistent to have a cardinal which has all the degrees of inaccessibility (describable in her notation) but no Mahlo cardinals at all.

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  • $\begingroup$ In my notation I use N to diagonalise over inaccessibility ($x^{th} x\text{-inaccessible}$), so the next ordinal $N+1$ denotes (1,0)-inaccesible, $N\cdot 2+1$ denotes (2,0)-inaccesible and so on (here it doesn't matter what $N$ if it is large enough) Then I define $N$ itself as the first ordinal which is inaccessible for every $a$ $\endgroup$
    – fejfo
    Commented May 26, 2018 at 7:49
  • $\begingroup$ My notation is similar to "Degrees of inaccessible cardinals... using meta-ordinals" in your paper. But I use $N$ instead of $\Omega$ and I always need $+1$ after the denationalization. What I ask is if the "limit" of this notation is $M$ $\endgroup$
    – fejfo
    Commented May 26, 2018 at 8:13
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    $\begingroup$ @fejfo The point is that the paper shows that the least cardinal having all of the degrees of inaccessibility (in the paper's notation, but I expect yours amounts to the same thing) can fail to be Mahlo. $\endgroup$ Commented May 26, 2018 at 11:13

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