Union and intersection of a regular and a non-regular language
Note that the languages $\emptyset$, $\{\epsilon\}$ and $\Sigma^*$ are regular. Let $L_2$ be any non-regular language over&nbsp;$\Sigma$.

Union. $\emptyset \cup L_2 = L_2$, which is non-regular; $\Sigma^*\cup L_2 = \Sigma^*$, which is regular.

Intersection. $\Sigma^*\cap L_2 = L_2$; $\emptyset\cap L_2 = \emptyset$.

Subtraction. $\Sigma^*\setminus L_2 = \overline{L_2}$, which is non-regular, since regular languages are closed under complementation.  That is, if $\overline{L_2}$ were regular, then $\overline{\overline{L_2}}=L_2$ would also have to be regular.  $\emptyset\setminus L_2 = \emptyset$, which is regular.

Concatenation. $\{\epsilon\}\cdot L_2 = L_2$; $\emptyset\cdot L_2 = \emptyset$.