Proving that $\bigcup_{M \in A} \{M_{ij}\} \cap \bigcup_{M \in B}\{M_{ij}\} = \bigcup_{M \in A \cap B} \{M_{ij}\}$ Let $A, B$ be two sets of $n \times n$ matrices, is it true that:
$$\forall i,j \in \{1,...,n\}: \bigcup_{M \in A} \{M_{ij}\} \cap \bigcup_{M \in B}\{M_{ij}\} = \bigcup_{M \in A \cap B} \{M_{ij}\}$$
I was able to prove that:
$$\bigcup_{M \in A \cap B} \{M_{ij}\} \subseteq \bigcup_{M \in A} \{M_{ij}\} \cap \bigcup_{M \in B}\{M_{ij}\}$$
But I'm having some trouble proving the opposite.
I started with: Let $i,j \in \{1,...,n\}$ and  $x \in \bigcup_{M \in A} \{M_{ij}\} \cap \bigcup_{M \in B}\{M_{ij}\}$, then:
$x \in \bigcup_{M \in A} \{M_{ij}\} \wedge x \in \bigcup_{M \in B}\{M_{ij}\}$ meaning that $\exists M \in A:M_{ij} = x$ and $\exists M' \in B: M'_{ij}=x$, but Now to colcude that $\exists K \in A \cap B: K_{ij}=x$ I would need that $M=M'$.
Is there a way to prove this or is this simply not true?
 A: As you suspected, the converse containment doesn't hold. To construct a simple counterexample, just take the sets $A$ and $B$ to be singleton sets with a matrix that has the same entry in $(i,j)$ spot and different entry in some other.
As an explicit counterexample, consider $A=\left\{\begin{bmatrix}0&0\\0&0\end{bmatrix}\right\}$ and $B=\left\{\begin{bmatrix}0&0\\0&1\end{bmatrix}\right\}$
Then $A\cap B=\emptyset$ and thus your LHS of the claim is the set $\{0\}$ for all $(i,j)$ except $(2,2)$ and the RHS is always $\emptyset$

A better counterexample:
To construct a counterexample for some $(i,j)$, make the sets $A,B$ such that $\bigcup_{x\in A} x_{ij}=\bigcup_{x\in B} x_{ij}$ but other entries are changed in the two matrices to make them not be in $A\cap B$
$$A=\left\{\begin{bmatrix}\color{red}0&0\\0&0\end{bmatrix}, \begin{bmatrix}\color{red}1&\underline{0}\\0&0\end{bmatrix}\right\}\\ B=\left\{\begin{bmatrix}\color{red}0&0\\0&0\end{bmatrix},\begin{bmatrix}\color{red}1&\underline{1}\\0&0\end{bmatrix}\right\}$$
Then, for $(i,j)=(1,1)$, the LHS of your claim is $\{\color{red}0,\color{red}1\}$ but the RHS of
your claim is $\{0\}$ because $A\cap B$ is only the singleton set with zero matrix.
To keep both sets of the claim non-empty, I let the zero matrix be in both $A$ and $B$ but changed the $(1,2)$-entry of the second element of $A,B$ keeping the $(1,1)$-entry same.
