Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
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

1 Answer 1

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.