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Why ($\operatorname{Con}_{FI}(A))$ is closed under arbitrary intersection?

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You'd better ask your question in the body, not in the title, MohammadSadegh. – Babak S. Dec 13 '12 at 7:07
ok my friend... & thanks alot for your attention to my question – MohammadSadegh YazdanParast Dec 15 '12 at 16:51
@You'r welcome,Mohammad. :) – Babak S. Dec 15 '12 at 16:53
up vote 2 down vote accepted

It’s very much like the proof that an arbitrary intersection of subgroups of a group is again a subgroup. Let $\Theta$ be any family of fully invariant congruences on $A$, and let $\theta=\bigcap\Theta$. Let $\sigma$ be any endomorphism of $A$, and let $a,b\in A$. Suppose that $a\,\theta\,b$. Then $a\,\rho\,b$ for all $\rho\in\Theta$, so $\sigma(a)\,\rho\,\sigma(b)$ for all $\rho\in\Theta$, and therefore $\sigma(a)\,\theta\,\sigma(b)$.

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thanks alot my friend. – MohammadSadegh YazdanParast Dec 15 '12 at 16:50
@Mohammad: You’re welcome. – Brian M. Scott Dec 15 '12 at 19:37

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