Proper class of inaccessibles and 1-inaccessible cardinals Hello I am starting to learn set theory and I am having some difficulty with these notions.
Let's say we have a 1-inaccessible cardinal which means it is stronger than the assertion that there is a proper class of inaccessibles.
How would this 1-inaccessible compare size wise to these cardinals, is it larger than every inaccessible in the proper class. I would assume so because if you cut the universe up to that cardinal there would be a proper class of inaccessibles but no 1-inaccessible
Thanks
 A: It seems like you're conflating size/outright implication with the more subtle notion of consistency strength.
A $1$-inaccessible is "at least as expensive" as a proper class of inaccessibles in the following sense: we can prove (in a very weak base theory) that if $\mathsf{ZFC}$ + "There is a $1$-inaccessible" is consistent then so is $\mathsf{ZFC}$ + "There is a proper class of inaccessibles." The proof is simple: given a model $M$ of the former theory, the related structure $(V_\kappa)^M$ where $\kappa$ is what $M$ thinks is the least $1$-inaccessible is a model of the latter theory.
HOWEVER, this does not mean that we have an outright implication between "There is a $1$-inaccessible" and "There is a proper class of inaccessibles," let alone that a $1$-inaccessible would somehow magically be bigger than a proper class of inaccessibles. For example, suppose $M$ were a model of $\mathsf{ZFC}$ + "There is a proper class of inaccessibles" + "There is a $1$-inaccessible." Letting $\alpha$ be (what $M$ thinks is) the smallest inaccessible above the least $1$-inaccessible, the related structure $(V_\alpha)^M$ satisfies "There is a $1$-inaccessible but not a proper class of inaccessibles."
A: Much of the large cardinal hierarchy is based on reflection. Taking a property, and asking "Okay, but what if we had a cardinal that was satisfying that property and also was a limit of cardinals with this property?"
This is very much in the spirit of understanding how countable ordinals work.
We have $\omega$, which is a limit ordinal. Great, case closed. No more successor ordinals, right? Well, not, there's $\omega+1,\omega+2,\dots$, then $\omega\cdot2$. Okay. So no more limit ordinals, right? Well, there's $\omega\cdot3,\omega\cdot4,\dots$ and even $\omega^2$. And each of those have their $\omega\cdot n+1$, $+2$, and so on. And we have $\omega^3,\omega^4$, etc.
But look how slowly we are going up the hierarchy of ordinals. This is painstakingly slow. It will take us forever before we even reach some reasonably large countable ordinal, let alone $\omega_1$. So we can define a hierarchy.

*

*First, look at limit ordinals. Good.

*Now look at limit ordinals that are limits of limit ordinals. Great.

*Now look at limit ordinals that are limits of limit ordinals that are themselves limit of limit ordinals. Fantastic.

*Keep going, each time reflecting more and more.

It will still take you forever to reach $\omega_1$ or any reasonably large countable ordinal. But we seem to get larger and larger with each step. We can also replace this by thinking about "closure under operations".

*

*Look at limit ordinals, these are closed under successors (so $\delta$ is limit if and only if for all $\alpha<\delta$, $\alpha+1<\delta$).

*Next, look at limit ordinals closed under addition, these are of the form $\omega^\gamma$ (this is ordinal exponentiation, and this corresponds to $\alpha,\beta<\delta$ if and only if $\alpha+\beta<\delta$).

*Next, look at limit ordinals that are closed under taking additively indecomposable ordinals (so, $\alpha<\delta$ if and only if the $\alpha$th indecomposable ordinal is below $\delta$).

*Keep going, each time taking closure under the previous class of ordinals.

Now we're going even faster than before. Still, we won't make it to $\omega_1$ in time for supper, or even to a reasonably very large countable ordinal. But the point remains. These ordinals get larger and larger, and each one is a limit of "the types that came before".
This does not negate the fact that there are a lot of countable ordinals which are not limit ordinals, or have the form $\alpha+\omega$, so they are not limit of limit ordinals, and so on.
But we can look at $V_\alpha$ and ask what kind of ordinals live there. In $V_\omega$, only finite ones. In $V_{\omega^2}$, only $\omega\cdot n+m$, so on.
This is kind of the same with inaccessible cardinals and their ilk.

*

*We first have regular cardinals.

*Next we have regular cardinals which are a limit of regular cardinals, these are the inaccessible cardinal.

*Then we ask for inaccessible cardinals which are limit of inaccessible cardinals, these are 1-inaccessible cardinals.

And so on.  At no point in the countable ordinals hierarchy we necessarily stopped having successor ordinals, or limits which are not-limit-of-limits, etc. So even if we have a 1-inaccessible cardinal, it does not preclude the possibility that there are larger inaccessible cardinals which are not 1-inaccessible cardinals.
Indeed, if there are exactly two 1-inaccessible cardinals, $\kappa<\lambda$, then each is the limit of inaccessible cardinals, so what can we say about those inaccessible cardinals in the interval $(\kappa,\lambda)$? They are not 1-inaccessible cardinals, that's for sure. But they are larger than $\kappa$, pretty much by definition.
What is confusing here, though, is that we treat $V_\kappa$ as a model of its own and ask for its properties. So if $\kappa$ is the least 1-inaccessible, then $V_\kappa$ has a proper class of inaccessible cardinals, but no 1-inaccessible cardinals. This means that the theory "$\sf ZFC+$ There is a 1-inaccessible cardinal" proves that there is a model—quite the nice model, too—of the theory "$\sf ZFC+$ There is a proper class of inaccessible cardinals", and therefore its consistency. Since no reasonable theory proves its own consistency, as per Gödel's second incompleteness theorem, this means that as far as consistency strength goes, 1-inaccessible cardinal is a much stronger axiom than "a proper class of inaccessible cardinals". Remember the slogan of large cardinals axioms: to prove more, you need to assume more.
We can generally compare large cardinals by their strength of compare their "smallest instances". So, the least inaccessible is smaller than the least 1-inaccessible; and the least 1-inaccessible is smaller than the least Mahlo cardinal; etc. But at some point things get weird, and some of the instances are not ordered the way you'd expect. But this generally happens somewhere pretty far in the large cardinal hierarchy (or at least pretty far from the basic notions of inaccessibility).
