Compact cardinal cannot be successor?

This is a follow-up question to $\kappa$ is compact $\implies$ $\kappa$ is regular. The definition I'm using for "compact" is the same as there.

I am trying to show if $\kappa$ is compact, then $\kappa$ cannot be a successor cardinal.

My first attempt was as follows: Assume $\kappa=\lambda^+$. Let $\{p_\alpha\}_{\alpha \in \kappa}$ be propositional variables, and let $p$ be $\bigvee_{\beta \in \lambda} p_\alpha$. Then the set of sentences $\Sigma=\{\neg p_\alpha\}_{\alpha \in \kappa} \cup \{p\}$ is obviously unsatisfiable, because if $p$ is true then at least some $p_\alpha$ must be true. Then, trying to reach the desired contradiction, I tried to show that any subset of $\Sigma$ having cardinality $< \kappa$ is satisfiable, but I realized this is false. (For example, $\{\neg p_\alpha\}_{\alpha \in \lambda} \cup \{p\}$ has size $\lambda < \kappa$ but it is not satisfiable.)

Perhaps there's a more creative way to pick $\Sigma$ which actually works?

Solution 1:

This strategy is taken from the hint in Jech, Exercise 17.17. Suppose $\kappa = \lambda^+$.

We work in the language with constant symbols $\{c_\alpha\mid \alpha\leq \kappa\}$, a binary relation $<$, and a ternary relation $R$. Consider the following theory $T$:

1. The sentence "$<$ is a linear order".
2. The schema $c_\alpha<c_\beta$ for all $\alpha<\beta\leq \kappa$.
3. The sentence "For all $x$, $R(x,y,z)$ is the graph of a function $f_x(y) = z$."
4. The sentence $\forall x\forall z(z<x\rightarrow \exists y\, R(x,y,z))$. This expresses that the range of $f_x$ contains the initial segment $\{z\mid z<x\}$.
5. The infinitary sentence $\forall x\forall y\forall z\, (R(x,y,z)\rightarrow \bigvee_{\alpha<\lambda} y = c_\alpha)$. This expresses that the domain of $f_x$ is contained in $\{c_\alpha\mid \alpha<\lambda\}$.

Now $T$ is inconsistent, since we must have $|\text{dom}(f_{c_\kappa})|\leq \lambda$ but $|\text{ran}(f_{c_\kappa})|\geq \kappa$ (since it contains $\{c_\alpha\mid \alpha < \kappa\}$).

Let $T'\subset T$ be a subset with $|T'|<\kappa$. Then $T'$ only contains $\lambda$-many of the sentences $c_\alpha<c_\beta$, so it only mentions at most $\lambda$-many of the constant symbols. Take $M$ to be the set of constant symbols mentioned in $T'$: the symbosl $\{c_\alpha\mid \alpha<\lambda\}$ mentioned in axiom (5), together with any of the constant symbols $c_\beta$ with $\beta\geq \lambda$ which are mentioned by instances of schema (2). We interpret the remaining constant symbols arbitrarily, say setting them all to be equal to $c_0$. We equip the model with its natural order, induced from the ordering on the indices of the constant symbols (its order type will be some ordinal $\xi$ with $\lambda \leq \xi < \kappa$). It remains to define the function $f_x$ for each element $x$. Note that $\{z\mid z<x\}$ has size $\leq \lambda$, since $|M| = \lambda$. So we can define $f_x$ to be an arbitrary function from $\{c_\alpha\mid \alpha<\lambda\}$ onto $\{z\mid z<x\}$ (unless $\{z\mid z<x\}$ is empty, in which case we can take $f_x$ to be the empty function).

We conclude that $L_{\kappa,\kappa}$ (stronger, $L_{\kappa,\omega}$) is not weakly compact.

Solution 2:

I'll show something stronger: If $L_{\kappa,\omega}$ is $\kappa$-compact, then $\kappa$ is a strong limit. The result of your previous question showed that $\kappa$ is regular, so we conclude that $\kappa$ is strongly inaccessible.

Assume $\kappa$ is not a strong limit, i.e., there exists $\lambda<\kappa$ such that $2^\lambda\geq \kappa$. Work in the language with proposition symbols $\{P_\alpha\mid \alpha<\lambda\}$, and consider the $L_{\kappa,\omega}$-theory $T$: $$\left\{\lnot\left(\bigwedge_{\alpha\in X}P_\alpha\land \bigwedge_{\alpha\notin X}\lnot P_\alpha\right)\,\middle\vert\, X\subseteq \lambda \right\}.$$ Note that each conjunction in $T$ is over a set of size $\leq \lambda < \kappa$. $T$ is unsatisfiable, because it explicitly rules out any assignment of truth values to the $P_\alpha$. But $|T| = 2^\lambda \geq \kappa$, so any subset of $T$ of size $<\kappa$ is a proper subset of $T$, which is satisfiable.

Of course, in the argument above, it is possible that $|T|>\kappa$, so the argument does not go through if we only assume $L_{\kappa,\omega}$ is weakly $\kappa$-compact (meaning that $\kappa$-compactness only applies to theories of size $\leq \kappa$).

If $L_{\kappa,\omega}$ is only weakly $\kappa$-compact, we can stil prove that $\kappa$ is a limit cardinal; hence, since $\kappa$ is regular, $\kappa$ is weakly inaccessible. My previous answer gave a proof of this following the hint given by Jech in Exercise 17.17 of Set Theory (Third Millennium Edition). Jonathan's answer shows that the idea can be dramatically simplified. And here's an even simpler version:

Assume $\kappa = \lambda^+$. Work in the language with constants $\{c_\alpha\mid \alpha<\kappa\}$ and $\{d_\gamma\mid \gamma<\lambda\}$. Consider the following $L_{\kappa,\omega}$-theory $T$: $$\{c_\alpha\neq c_\beta\mid \alpha< \beta<\kappa\}\cup \{\forall x\,\bigvee_{\gamma<\lambda} x = d_\gamma\}$$ $T$ is not satisfiable, because the schema $c_\alpha\neq c_\beta$ implies that the size of a model is $\geq \kappa$, while the sentence $\forall x\,\bigvee_{\gamma<\lambda} x = d_\gamma$ ensures that the size of a model is $\leq \lambda$. But any subset $T'\subseteq T$ with $|T'|<\kappa$ only asserts that at most $\lambda$-many of the constants $c_\alpha$ are distinct, so $T'$ admits a model of cardinality $\lambda$.

So the natural question is: does weak $\kappa$-compactness of $L_{\kappa,\kappa}$ imply that $\kappa$ is a strong limit (and hence strongly inaccessible?). I didn't know the answer, but according to the comments to this MathOverflow question, the answer is no. Apparently the result is due to Solovay and Kunen, and a proof is given in the paper Infinitary compactness without strong inaccessibility by William Boos.

Note that many sources define "weakly compact cardinal" in such a way that weakly compact cardinals are strongly inaccessible. For example, Jech says a cardinal $\kappa$ is weakly compact if it is uncountable and $\kappa\to (\kappa)^2$, and this definition implies that $\kappa$ is a strong limit. Then he proves (Theorem 17.13) that if $\kappa$ is (strongly) inaccessible and the logic $L_{\kappa,\omega}$ is weakly compact, then $\kappa$ is weakly compact.

Solution 3:

Assume $\kappa = \lambda^+$.

Consider a language which has constants $\langle c_\alpha : \alpha < \kappa \rangle$ and one function symbol $f$. Let $T$ be a theory containing

• A sentence expressing that $f$ is a bijection: $$\forall y \exists! x (f(x) = y),$$
• for each $\alpha < \kappa$, a sentence expressing that $f^{-1}(c_\alpha) = c_\beta$ for some $\beta < \lambda$. Namely, $$ \forall x (f(x) = c_\alpha \rightarrow \bigvee_{\beta < \lambda} x = c_\beta ). $$
• for each $\alpha, \beta < \kappa$, a sentence expressing that $c_\alpha \neq c_\beta$.

It is clear that this is unsatisfiable since in any model $f$ restricted to the first $\lambda$ many constants induces a bijection between $\lambda$ and $\kappa = \lambda^+$. But any subtheory of size $< \kappa$, i.e. of size $\leq \lambda$, mentions at most $\lambda$ many constants. To get a model it suffices to use a bijection between the first $\lambda$ many coordinates and $\lambda$ many constants including those that appear in our subtheory.