I was messing around with the intuition behind the size of weakly compact cardinals (in their usual characterization). I found an interesting, seemingly weaker LCA which still implies weak inaccessibility.

I started by making an intuitively powerful property that can be stated as an $\mathcal{L}_{\kappa,\kappa}$ scheme. I ended up with three properties:

- $\text{LZ}_\lambda$ for $\lambda<\kappa$ is the $\mathcal{L}_{\kappa,\kappa}$-sentence "$\exists_{\alpha<\lambda}x_\alpha(\bigwedge_{\alpha<\lambda}\bigwedge_{\beta\neq\alpha}x_\alpha\neq x_\beta)$" which is actually true in a structure $\mathcal{M}$ iff it's universe has size at least $\lambda$ (it's not hard to see why).
- $\text{LA}_\lambda$ for $\lambda<\kappa$ is the $\mathcal{L}_{\kappa,\kappa}$-sentence "$\exists_{\alpha<\lambda}x_\alpha\forall X(\bigvee_{\alpha<\lambda}x_\alpha=X)$" which is true in a structure $\mathcal{M}$ iff it's universe has size
**at most**$\lambda$.

So with $\mathcal{L}_{\kappa,\kappa}$ we can actually bound the size of our structure from above and below to cardinals $<$$\kappa$. It's a strong failure of the generalized Lowenheim-Skolem below $\kappa$.

## Applying it to Compactness

If $\kappa$ is weakly compact, and $T$ is an $\mathcal{L}_{\kappa,\kappa}$-theory with at most $\kappa$-many symbols, and $T$ is satisfiable, then $T+S$ for a scheme $S=\{A_0,A_1...A_{n<\omega}...\}$ is satisfiable if and only if $T+A_0$ is satisfiable and $T+A_0+A_1$ is satisfiable (etc.)

We can use this to our power. Let $\text{LS}_\kappa$ be the $\mathcal{L}_{\kappa,\kappa}$-scheme which has $\text{LZ}_\lambda$ for all $\lambda<\kappa$. Clearly, $\mathcal{M}\models\text{LS}_\kappa$ iff $|M|\geq\kappa$.

However, because $\kappa$ is weakly compact, $T+\text{LS}_\kappa$ is satisfiable (there is a model of $T$ of size at least $\kappa$) if and only if $T+\text{LZ}_\lambda$ is satisfiable for every $\lambda<\kappa$. However, that is only true if and only if for any $\lambda<\kappa$, there is a model of $T$ larger than $\lambda$. In other words, if there are arbitrarily large models of $T$ below $\kappa$, there is a model $\mathcal{M}\models T$ with $|M|\geq\kappa$.

**Therefore, if $\kappa$ is weakly compact and $T$ is an $\mathcal{L}_{\kappa,\kappa}$-theory of at most $\kappa$-many symbols with arbitrarily large models of $T$ below $\kappa$, there is a model $\mathcal{M}\models T$ with $|M|\geq\kappa$.**

A similar argument shows that if $\kappa$ is strongly compact and $T$ is ANY $\mathcal{L}_{\kappa,\kappa}$-theory with arbitrarily large models of $T$ below $\kappa$, there is a model $\mathcal{M}\models T$ with $|M|\geq\kappa$.

Furthermore, if $\kappa$ is extendible and $T$ is any $\mathcal{L}^n_{\kappa,\kappa}$-theory for finite $n$ with arbitrarily large models of $T$ below $\kappa$, there is a model $\mathcal{M}\models T$ with $|M|\geq\kappa$.

## Turing Into a Unique Property

This property makes $\kappa$ intuitively quite large. If you look at the "shadow" of $\kappa$ left by $\aleph_0$ (because remember, this also applies to $\aleph_0$) then you find that any first-order finitary theory $T$ with arbitrarily large finite models must also have an infinite model.

Intuitively, this is saying that "first-order finitary logic cannot talk about infinite models very well, but it sure can talk about finite models." The generalization to this new property is that "$\mathcal{L}_{\kappa,\kappa}$ cannot talk about $\geq$$\kappa$-sized models very well, but it sure can talk about $<$$\kappa$-sized models."

Hey, but wait! We haven't named this property yet! Let's do that.

## Definitions

Let $T$ be a theory. $T$ is **$\kappa$-unboundedly satisfiable** iff for any $\lambda<\kappa$, there is a model of $T$ of size at least $\lambda$.

Let $\kappa$ be **weakly Skolem** iff $\kappa$ is uncountable and regular and every $\kappa$-unboundedly satisfiable $\mathcal{L}_{\kappa,\kappa}$-theory $T$ of at most $\kappa$-many symbols has a model of size at least $\kappa$.

Let $\kappa$ then be **Skolem** iff every $\kappa$-unboundedly satisfiable $\mathcal{L}_{\kappa,\kappa}$-theory $T$ has a model of size at least $\kappa$ (no restrictions on the symbols of $T$).

Let $\kappa$ then be **Extendibly Skolem** iff every $\kappa$-unboundedly satisfiable $\mathcal{L}^n_{\kappa,\kappa}$-theory $T$ has a model of size at least $\kappa$.

Note - every weakly compact cardinal is weakly Skolem, every strongly compact cardinal is Skolem, and every extendible cardinal is extendibly Skolem. Every extendibly Skolem cardinal is Skolem, and every Skolem cardinal is weakly Skolem.

## Proof of Weak Inaccessibility

To show that every weakly Skolem cardinal is inaccessible, I just have to show that it is a limit cardinal. This is easy enough.

Let $\kappa=\lambda^+$ be weakly Skolem. Then, $\text{LA}_\lambda$ is an $\mathcal{L}_{\kappa,\kappa}$-sentence. Clearly, $\lambda\models\text{LA}_\lambda$, and for any $\mu<\kappa$, $\mu\leq\lambda$. Therefore, there are arbitrarily large models of $\text{LA}_\lambda$ below $\kappa$.

This would imply there is a model $\mathcal{M}$ of $\text{LA}_\lambda$ of size at least $\kappa$, which is impossible by definition of $\text{LA}_\lambda$.

## The Questions

- Are these cardinals strongly inaccessible? If not, are they equivalent to weak inaccessibility?
- Is weak Skolem-ness equivalent to weak compactness? Similarly with Skolem-ness and extendible Skolem-ness
- What properties do these cardinals have?

EDIT: By User Yair Hayut, weak Skolem-ness need not imply strong inaccessibility. However, any strongly inaccessible weakly Skolem cardinal is weakly compact.

at least$\kappa $. $\endgroup$ – Keith Millar Sep 5 '18 at 14:39