Is the Mahlo ordinal the first cardinal unreachable using inaccessibility and diagonalisation?

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" $$a_0(x)=x+1$$ $$a_\alpha(x)=\text{the x^{th} ordinal in } \{y\mid\gamma $$\text{if }\operatorname{cf}(\alpha) $$a_\alpha(x)=a_{\alpha[x]}(x)\text{ if }\operatorname{cf}(\alpha)=N$$ $$N=\min\{y\mid \gamma

$$N$$ 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 $$M$$

• You define $N$ in terms of itself. Commented May 25, 2018 at 16:54
• 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 $2x=x+1$ )? Commented May 25, 2018 at 17:01
• JDH explains why the answer is a definitive "no" in his answer to this question: math.stackexchange.com/questions/24505/… Commented Jun 5, 2018 at 8:44

• 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 $a$ Commented May 26, 2018 at 7:49
• 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 $M$ Commented May 26, 2018 at 8:13