[proofplan] We verify the three defining properties of an [equivalence relation](/page/Equivalence%20Relation). Reflexivity follows because every indexed family agrees with itself on all of $I$, and $I$ belongs to every filter on $I$. Symmetry follows because equality in each coordinate is symmetric, so the agreement set for $a$ with $b$ is the same as the agreement set for $b$ with $a$. Transitivity follows by intersecting the two large agreement sets and using the upward-closure axiom for filters. [/proofplan] [step:Record the definition of large agreement modulo the filter] For elements $a = (a_i)_{i \in I}$ and $b = (b_i)_{i \in I}$ of $\prod_{i \in I} M_i$, define the agreement set $E(a,b) \subset I$ by \begin{align*} E(a,b) := \{i \in I : a_i = b_i\}. \end{align*} By definition of large agreement modulo $\mathcal F$, we have \begin{align*} a \sim_{\mathcal F} b \quad \iff \quad E(a,b) \in \mathcal F. \end{align*} The rest of the proof verifies the three equivalence-relation axioms using this definition. [/step] [step:Show that every element agrees with itself on a set in the filter] Let $a = (a_i)_{i \in I} \in \prod_{i \in I} M_i$. Define the self-agreement set $A \subset I$ by \begin{align*} A := \{i \in I : a_i = a_i\}. \end{align*} Since equality is reflexive in each set $M_i$, we have $A = I$. Because $\mathcal F$ is a filter on $I$, it contains $I$. Hence $A \in \mathcal F$, so $a \sim_{\mathcal F} a$. Therefore $\sim_{\mathcal F}$ is reflexive. [/step] [step:Use symmetry of coordinate equality to reverse large agreement] Let $a = (a_i)_{i \in I}$ and $b = (b_i)_{i \in I}$ be elements of $\prod_{i \in I} M_i$, and suppose $a \sim_{\mathcal F} b$. Define the agreement sets $A, B \subset I$ by \begin{align*} A &:= \{i \in I : a_i = b_i\}, \\ B &:= \{i \in I : b_i = a_i\}. \end{align*} For each $i \in I$, the statements $a_i = b_i$ and $b_i = a_i$ are equivalent by symmetry of equality in $M_i$. Hence $A = B$. Since $a \sim_{\mathcal F} b$, we have $A \in \mathcal F$, and therefore $B \in \mathcal F$. Thus $b \sim_{\mathcal F} a$. Hence $\sim_{\mathcal F}$ is symmetric. [/step] [step:Intersect two large agreement sets to prove transitivity] Let $a = (a_i)_{i \in I}$, $b = (b_i)_{i \in I}$, and $c = (c_i)_{i \in I}$ be elements of $\prod_{i \in I} M_i$. Suppose $a \sim_{\mathcal F} b$ and $b \sim_{\mathcal F} c$. Define the agreement sets $A, B, C \subset I$ by \begin{align*} A &:= \{i \in I : a_i = b_i\}, \\ B &:= \{i \in I : b_i = c_i\}, \\ C &:= \{i \in I : a_i = c_i\}. \end{align*} By the hypotheses, $A \in \mathcal F$ and $B \in \mathcal F$. Since $\mathcal F$ is closed under finite intersections, $A \cap B \in \mathcal F$. For every $i \in A \cap B$, we have $a_i = b_i$ and $b_i = c_i$, so by transitivity of equality in $M_i$, $a_i = c_i$. Therefore $A \cap B \subset C$. Since $\mathcal F$ is upward closed and $A \cap B \in \mathcal F$, it follows that $C \in \mathcal F$. Hence $a \sim_{\mathcal F} c$. Thus $\sim_{\mathcal F}$ is transitive. [/step] [step:Conclude that large agreement modulo the filter is an equivalence relation] We have shown that $\sim_{\mathcal F}$ is reflexive, symmetric, and transitive on $\prod_{i \in I} M_i$. Therefore $\sim_{\mathcal F}$ is an equivalence relation on $\prod_{i \in I} M_i$. [/step]