Does the following define a Mahlo cardinal?

Question

Let M be a cardinal with the following properties:
- M is regular
- $\kappa < M \implies 2^\kappa < M$
- $\kappa < M \implies s(\kappa) < M$ where $s(\kappa)$ is the smallest strongly inaccessible cardinal strictly greater than $\kappa$

My question is: Is M a Mahlo cardinal ? If so, how does the definition above connect to the usual definition in terms of stationary sets ?

Motivation: My intuition about a Mahlo cardinal is that it you cannot reach it by taking unions, power sets or "the next inaccessible cardinal", which is my definition above. My worry is, I might have arrived at something much smaller than Mahlo.

Answer 1

No, your condition doesn't imply Mahloness.

First, note that your first two conditions simply state that $M$ is inaccessible, and the third one gives that $M$ is limit of inaccessibles. Now consider the first cardinal $M_0$ which is inaccessible limit of inaccessibles. That is, every inaccessible below $M_0$ is not a limit of inaccessibles. Now consider $C$ consisting of limits of the set $I$ of inaccessible cardinals smaller than $M_0$. By choice of $M_0$, $C$ contains no inaccessibles below $M_0$, so if we show $C$ is a club below $M_0$, we will get that $M_0$ is not Mahlo.

$C$ is closed: this is true because the set of limit points of any set is closed. For the sake of completion, I will sketch the argument: suppose that each of $\alpha_i$ is a limit point of $I$, where $(\alpha_i)_{i<\kappa}$ is an increasing sequence of cardinals with limit $A$. Let $\beta<A$. This means that there is some $\alpha_i$ between $\beta$ and $A$. Since $\beta<\alpha_i$, there is some element of $I$ which is between $\beta$ and $\alpha_i$, hence between $\beta$ and $A$. So $A$ is a limit point of $I$, so $A\in C$. Hence $C$ is closed.

$C$ is unbounded below $M_0$: Let $\alpha<M_0$. Then $s(\alpha),s(s(\alpha)),s(s(s(\alpha))),...$ is an increasing sequence in $I$ of length $\omega$. Since cofinality of $M_0$ is (clearly) greater than $\omega$, limit of this sequence, which is an element of $C$, is between $\alpha$ and $M_0$. Hence $C$ is unbounded below $M_0$.

Indeed, the condition "inaccessible limit of inaccessibles" (or simply "regular limit of inaccessibles") is a condition known as 1-inaccessibility. It can be generalized: we can speak of 2-inaccessibles, which are regular limits of 1-inaccessibles, 3-inaccessibles which are regular limits of 2-inaccessibles and so on. This can be obviously extended to a successor ordinal, and for limit ordinal $\alpha$ (e.g. $\alpha=\omega$) we say that cardinal is $\alpha$-inaccessible if it's $\beta$-inacessible for $\beta<\alpha$ (this isn't circular definition, but rather it proceeds by transfinite induction).

For every ordinal $\alpha$, notion of $\alpha+1$-inaccessibility is strictly stronger than that of $\alpha$-inaccessibility. As can be shown, every Mahlo cardinal $\kappa$ is not only 1-inaccessible, 2-inaccessible, $\omega$-inaccessible, but even $\kappa$-inaccessible. Notion of Mahloness is even stronger than that, but for that, I'd recommend reading this section on Wikipedia article.

Answer 2

Wojowu has answered the question, but it might be useful to record here why the first $M$ that satisfies your conditions is not a Mahlo cardinal. Consider the set $C$ of those cardinals $\lambda<M$ that satisfy $(\forall\kappa<\lambda)\,2^\kappa<\lambda$ and $(\forall\kappa<\lambda)\,s(\kappa)<\lambda$, i.e., $\lambda$ satisfies the second and third of your hypotheses about $M$. (Actually, the third hypothesis subsumes the second, because $s(\kappa)>2^\kappa$.) Note that none of these $\lambda$'s is regular, because I assumed that $M$ is the first cardinal satisfying all three of your conditions. Now suppose, toward a contradiction, that $M$ were Mahlo. Then $C$ could not be closed and unbounded in $M$, because the definition of Mahlo says that every closed unbounded subset of $M$ contains a regular cardinal.

It's easy to check that $C$ is closed, so it would have to be bounded, i.e., the supremum $\sigma$ of $C$ would be $<M$ (and would therefore be an element of $C$ as $C$ is closed). Now define a countable increasing sequence of cardinals $\sigma_n<M$ by $\sigma_0=\sigma$ and $\sigma_{n+1}=s(\sigma_n)$. The supremum $\tau$ of this sequence satisfies the second and third of your hypotheses, so, if it were $<M$, then it would be in $C$, contrary to $\sigma$ being the supremum of $C$. So $\tau=M$. But this is absurd, as $\tau$ has countable cofinality whereas $M$ is uncountable and regular.

That completes the proof that the first $M$ satisfying your conditions isn't Mahlo, but let me add a remark that might help clarify what Mahlo cardinals are "about". Note that the proof above didn't use very much about the functions $\kappa\mapsto 2^\kappa$ and $s$ that occur in your second and third hypotheses. You could add more conditions of the form "$(\forall\kappa<\lambda)\,f(\kappa)<\lambda$" for other functions $f$, and the same argument will still work. The first regular cardinal with any specified closure properties of this sort is not Mahlo. (You could even take $f(\kappa)$ to be the first Mahlo cardinal above $\kappa$; then the first $M$ would be an inaccessible limit of Mahlo cardinals, but would not itself be Mahlo.) The key idea that the definition of Mahlo cardinals tries to capture is the idea of being unobtainable (from below) by any sort of closure properties.

Answer 3

Thanks a lot for your answers, Wojowu and Andreas. I find I am not able to simply reply with a comment while being a guest, so I am writing it this way.

The mental picture I am getting of Mahlo versus inaccessibles is somewhat like $\aleph_1$ versus the Veblen hierarchy of countable ordinals - you keep recursively strengthening definitions and taking fixed points, but just never get there. Don't know how strong the analogy is, though.