Aleph 0 as a large cardinal

This probably isn't the kind of thing anyone just knows off hand, so anyone who's going to answer the question is just going to look at a list of large cardinal axioms and their definitions, and try to see which ones are satisfied by $\aleph _0$ and which aren't. You could've probably done this just as well as I could have, but I decided I'd do it just for the heck of it.

First of all, this doesn't cover all large cardinal axioms. Second, many large cardinal axioms have different formulations which end up being equivalent for uncountable cardinals, or perhaps inaccessible cardinals, but may end up inequivalent in the context of $\aleph _0$. So even for the large cardinals that I'll look at, I might not look at all possible formulations.

1. weakly inaccessible - yes, obviously
2. inaccessible - yes, obviously
3. Mahlo - no, since the only finite "inaccessibles" are 0, 1, and 2 as noted by Michael Hardy in the comments, and the only stationary subsets of $\omega$ are the cofinite ones
4. weakly compact:
   • in the sense of the Weak Compactness Theorem - yes, by the Compactness Theorem
   • in the sense of being inaccessible and having the tree property - yes, by Konig's Lemma
   • in the sense of $\Pi ^1 _1$-indescribability - no, it's not even $\Pi ^0 _2$-indescribable as witnessed by the sentence $\forall x \exists y (x \in y)$
5. indescribable - no, since it's not even $\Pi ^0 _2$-indescribable
6. Jonsson - no, the algebra $(\omega, n \mapsto n \dot{-} 1)$ has no proper infinite subalgebra
7. Ramsey - no, the function $F : [\omega ]^{< \omega} \to \omega$ defined by $F(x) = 1$ if $|x| \in x$ and $0$ otherwise has no infinite homogeneous set
8. measurable:
   • in the sense of ultrafilters - yes, by Zorn's Lemma, and because filters are $\omega$-complete by definition, i.e. closed under finite intersections
   • in the sense of elementary embeddings - no, obviously
9. strong - no, obviously (taking the elementary embedding definition)
10. Woodin - ditto
11. strongly compact:
    • in the sense of the Compactness Theorem - yes, by the Compactness Theorem
    • in the sense of complete ultrafilters - yes, as in the case of measurables
    • in the sense of fine measures - yes, by Zorn's Lemma, and because filters are $\omega$-complete by definition
12. supercompact:
    • in the sense of normal measures - no, if $x \subset \lambda$ is finite and $X$ is the collection of all finite subsets of $\lambda$ which contain $x$, then the function $f : X \to \lambda$ defined by $f(y) = \max (y)$ is regressive, but for any $Y \subset X$, if $f$ is constant on $Y$ with value $\alpha$, then Y avoids the collection of finite subsets of $\lambda$ which contain $\{ \alpha + 1\}$ and hence Y cannot belong to any normal measure on $P_{\omega }(\lambda)$
    • in the sense of elementary embeddings - no, obviously
13. Vopenka - no, take models of the empty language of different (finite) sizes
14. huge - no, obviously (taking the elementary embedding definition)

There is a sense in which rank-into-rank axioms could be considered generilzations of $\omega$. Woodin discussed this in his paper Suitable Extender Models $I$, given the following conjecture, which I denote $(^*)$.

$(^*)$ Assume $V=\text{Ultimate}-L$. Then there is some $j: V_{\lambda+1}\prec V_{\lambda+1}$ if and only if $L(P(\lambda))\nvDash AC$.

In the same vein is the following conjecture, described by Vincenzo Dimonte in $I0$ and rank-into-rank axioms.

Question 11.2: Is it true that $\text{Ultimate}-L\vDash I0(\lambda)$ if and only if $\text{Ultimate}-L\vDash L(V_{\lambda+1})\nvDash AC$?

If either one of these were true, then it would be a shocking affirmation of the existence of $I1$ and $I0$ cardinals respectively (Assuming the proof does not show that there are none). The reason for this is that if $V=\text{Ultimate}-L$, then there is a proper class of Woodin cardinals and so in particular $L(\mathbb R)=L(P(\omega))=L(V_{\omega+1})\vDash AD$ and so $L(P(\omega))=L(V_{\omega+1})\nvDash AC$.

This corresponds with the sense in which it is not the critical point $\kappa$ of $j: V_{\lambda+1}\prec V_{\lambda+1}$ which is important, but the $\lambda$ itself. Note that $\lambda$ is a strong limit of cofinality $\omega$, and also $\omega$ is a strong limit of cofinality $\omega$, and finally a theorem of Shelah is that if $\lambda$ is a strong limit of uncountable cofinality, then $L(P(\lambda))\vDash AC$.