# Let $\mathcal A$ be a family of equivalence relations on $X$. Prove that $\bigcap_{\mathcal{R}\in\mathcal{A}} \mathcal R$ is an equivalence relation

Let $\mathcal A$ be a family of equivalence relations on $X$. Prove that $\bigcap_{\mathcal{R}\in\mathcal{A}}\mathcal{R}$ is an equivalence relation

Let's take an arbitrary ordered pair $(x,y)$. If this pair wants to be a member of $\bigcap_{\mathcal{R}\in\mathcal{A}}\mathcal{R}$, then for all subsets $S$ of $\mathcal A$, that is - equivalence relation, the ordered pair must be in the relation, and so, in more formal terms:
$$(x,y) \in \bigcap_{\mathcal{S}\in\mathcal{A}}\mathcal{S} \iff(\forall S \in \mathcal A)(xSy)$$
Now, we need to check if this new relation is:

(1) Reflexive
This relation is in fact reflexive, because all $S$ relations are reflexive.

(2) Symmetric
Once again - it seems obvious, because all of $S$ relations are symmetric.

(3) Transitive
$$(x,y) \in \bigcap \mathcal A \land (y,z) \in \bigcap \mathcal A \iff (\forall S\subseteq\mathcal A)(xSy \land ySz) \Rightarrow \\ \Rightarrow (\forall S \subseteq \mathcal A)(xSz) \iff (x,z) \in \bigcap \mathcal A$$

And so this relation is a relation of equivalence.
Is my solution correct? This just seems too easy.

• Perfectly correct. Dec 11, 2017 at 18:45

And there is in fact a small error: when you wrote $\forall S\subseteq A$, what you meant was $\forall S\in A$.