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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.

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  • $\begingroup$ Perfectly correct. $\endgroup$
    – amsmath
    Dec 11, 2017 at 18:45

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In my opinion, your answer is inconsistent. Transitivity isn't less obvious that reflexivity and symmetry. You should prove each property with the same level of detail. Actually, for the first two properties, what you do amounts to saying “it is true because it is true”.

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

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  • $\begingroup$ I spotted the error, too. But OP corrected it in the first line, so I thought they would have done it everywhere... $\endgroup$
    – amsmath
    Dec 11, 2017 at 20:09

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