# My question is why the set of fully invariant congruence of a set A is closed under arbitrary intersection?

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

-
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

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 $\sigma(a)\,\theta\,\sigma(b)$.

-
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