5. The proof of the subdirect representation theorem

5.1 Lemma. Given $a \neq b$ in ${\cal A}$, there exists a congruence relation $\theta$ maximal with respect to the property $a \not \equiv b \; (\theta)$.

Proof. Let ${\cal S} = \{\theta \in$   Con$({\cal A}) : \langle a,b \rangle \not \in \theta\}$. Then ${\cal S}$ is not empty, since $0 \in {\cal S}$. Suppose ${\cal C}$ is a chain of members of ${\cal S}$, where each relation is regarded as a subset of ${\cal A} \times {\cal A}$. Then $\bigcup _ {\theta \in {\cal C}} \theta \in {\cal S}$, since all aspects of being in ${\cal S}$ (specifically, being an equivalence relation, being compatible with the operations of ${\cal A}$, and not containing $\langle a,b \rangle$) can be checked using finitely many elements at a time and so can be checked inside just one member of ${\cal C}$ at a time. Then by Zorn's Lemma, ${\cal S}$ has a maximal member. $\Box$

Let $\theta _ {ab}$ be one such congruence relation maximal with respect to not identifying $a$ and $b$. Here $\theta _ {ab}$ is in contrast to con$(a,b)$, the smallest congruence relation that identifies $a$ and $b$. In fact, $\theta _ {ab}$ can be described as a $\theta$ maximal with respect to the property $\theta \not \geq$   con$(a,b)$.

5.2 Observation. For $a \neq b$ in ${\cal A}$, in Con$({\cal A})$ there is a least element $> \theta _ {ab}$, namely $\theta _ {ab} \vee$   con$(a,b)$.

5.3 Observation. ${\cal A}/\theta _ {ab}$ is subdirectly irreducible. Indeed, by Observation 1 and the Correspondence Theorem, Con$({\cal A}/\theta _ {ab})$ has a least element $> 0$ and so is subdirectly irreducible.

5.4 Observation. $\bigcap _ {a \neq b} \theta _ {ab} = 0$ in Con$({\cal A})$, where $a,b$ range over ${\cal A}$.

Proof of the Representation Theorem. By Observation we have ${\cal A} \hookrightarrow \prod _ {a \neq b} {\cal A}/\theta _ {ab}$, and by Observation each ${\cal A}/\theta _ {ab}$ is subdirectly irreducible.