For two classes $K_1,K_2$ of similar type, show $P_U(K_1 \cup K_2) = P_U(K_1) \cup P_U(K_2)$

Question

For two classes $K_1,K_2$ of similar type, show $P_U(K_1 \cup K_2) = P_U(K_1) \cup P_U(K_2)$. One direction seems fairly straight forward.

Namely, $P_U(K_1) \subseteq P_U(K_1 \cup K_2)$ and $ P_U(K_2) \subseteq P_U(K_1 \cup K_2)$. We can see this because an ultraproduct B of elements from $K_j, j = 1,2$ will be an ultraproduct of elements of $K_1 \cup K_2$ since $K_j \subseteq K_1 \cup K_2, j = 1,2$. So $ P_U(K_1) \cup P_U(K_2) \subseteq P_U(K_1 \cup K_2) $.

For the converse it is not so straightforward. Let $\prod_{i \in I }A_i$ where $U$ is an ultrafilter over $I$. I was told to consider sets $J_1 = \{i \in I : A_i \in K_1\}$ and $J_2 = \{i \in I : A_i \in K_2\}$.

It was suggested that $J_1 \cup J_2 = I$ and I am not seeing why exactly. Further it was mentioned that since $U$ is an ultrafilter that $J_1 \in U$ or $J_2 \in U$. I am not sure how to use this information to get the desired result.

Explanation as to why these sets are important and how to use them is appreciated.

Answer 1

Let $\prod_{i \in I} A_i / U \in P_U(K_1 \cup K_2)$, where $U$ is some ultrafilter over $I$. Consider the sets $J_1$ and $J_2$ you've defined. Take any $i \in I$, we have $A_i \in K_1$ or $A_i \in K_2$ since we are considering the ultraproduct of the members of $K_1 \cup K_2$. It means that $i \in J_1$ or $i \in J_2$, i.e. $I = J_1 \cup J_2$.

Now, since $U$ is an ultrafilter we have that $J_1 \in U$ or $J_2 \in U$. Let's assume that $J_1 \in U$. Since for $i \in J_1$ we have $A_i \in K_1$ the idea is to somehow construct an ultraproduct over $J_1$ (it will belong to the class $P_U(K_1)$) and to relate it to our initial ultraproduct. Let's do this!

Consider the following set $U_{J_1} = \{X \cap J_1 \mid X \in U\}$. You can check that $U_{J_1}$ is a filter over $J_1$ (here we are using $J_1 \in U$). I claim that $U_{J_1}$ is actually an ultrafilter over $J_1$. Suppose we have $A \subseteq J_1$ with $A \notin U_{J_1}$. Since $A = A \cap J_1 \notin U_{J_1}$ it must be the case that $A \notin U$ and hence we have $I \setminus A \in U$. But then (since $J_1 \subseteq I$) we have $J_1 \setminus A = (I \setminus A) \cap J_1 \in U_{J_1}$.

Now I claim that $\prod_{i \in I} A_i / U \cong \prod_{j \in J_1} A_j / U_{J_1} \in P_U(K_1)$. Consider the following surjective homomorphisms:

1. $\varphi$ and $\psi$ are natural projections onto the quotients $$\varphi \colon \prod_{i \in I} A_i \to \prod_{i \in I} A_i / U, \quad \psi \colon \prod_{j \in J_1} A_j \to \prod_{j \in J_1} A_j / U_{J_1}.$$
2. $\pi$ is the restriction homomorphism ($f \mapsto f_{J_1}$) $$\pi \colon \prod_{i \in I} A_i \to \prod_{j \in J_1} A_j.$$

The composition $\psi\pi$ is also surjective. By the homomorphism theorem we have $$\prod_{i \in I} A_i / U \cong \prod_{i \in I} A_i / \mathrm{ker}\,\varphi, \quad \prod_{j \in J_1} A_j / U_{J_1} \cong \prod_{i \in I} A_i / \mathrm{ker}\,\psi\pi.$$

It is left to show that $\mathrm{ker}\,\varphi = \mathrm{ker}\,\psi\pi$. For any $f, g \in \prod_{i \in I} A_i$ we have the following chain of equivalences (using the definition of ultraproduct and $U_{J_1}$): $$f (\mathrm{ker}\,\psi\pi) g \Leftrightarrow J_1 \cap \{i \in I \mid f(i) = g(i)\} = \{j \in J_1 \mid f_{J_1}(j) = g_{J_1}(j)\} \in U_{J_1}\\ \Leftrightarrow \{i \in I \mid f(i) = g(i)\} \in U \Leftrightarrow f(\mathrm{ker}\,\varphi)g.$$

Thus $\mathrm{ker}\,\varphi = \mathrm{ker}\,\psi\pi$ and $$\prod_{i \in I} A_i / U \cong \prod_{i \in I} A_i / \mathrm{ker}\,\varphi = \prod_{i \in I} A_i / \mathrm{ker}\,\psi\pi \cong \prod_{j \in J_1} A_j / U_{J_1}.$$