operations which are closest under regularity - other seeing

Question

We have: $L_1, L_2, $ regular and $L_3$ irreguar. Now: $L_1\cap L_2$ is regular.
$L_1\cap L_4 = L_3$
Can I say that $L_4$ is irregular ? The same question about $\cdot, \cup$

Answer 1

The class of regular languages is closed under intersection, so if $L_4$ were a regular language, then $L_1\cap L_4=L_3$ would be regular. But $L_3$ is not regular, so $L_4$ cannot be regular. The class of regular languages is also closed under union and concatenation, so you can argue in exactly the same way about $L_1\cup L_4$ and $L_1L_4$.