are NP Complete languages closed under any regular operations?
For all of the examples in this answer, I'm taking the alphabet to be $\{0,1\}$. Note that the languages $\emptyset$ and $\{0,1\}^*$ are definitely not NP-complete.


The class of NP-complete languages is not closed under intersection.  For any NP-complete language&nbsp;$L$, let $L_0 = \{0w\mid w\in L\}$ and $L_1 = \{1w\mid w\in L\}$.  $L_0$ and&nbsp;$L_1$ are both NP-complete but $L_0\cap L_1 = \emptyset$.
The class of NP-complete languages is not closed under union. Given the NP-complete languages $L_0$ and&nbsp;$L_1$ from the previous part, let $L'_0 = L_0 \cup \{1w\mid w\in \{0,1\}^*\}\cup\{\varepsilon\}$ and $L'_1 = L_1\cup \{0w\mid w\in \{0,1\}^*\}\cup\{\varepsilon\}$.  $L'_0$ and&nbsp;$L'_1$ are both NP-complete but $L'_0\cup L'_1 = \{0,1\}^*\!$.
The class of NP-complete languages is not closed under concatenation. Consider the NP-complete languages $L'_0$ and&nbsp;$L'_1$ from the previous part.  Since both languages contain&nbsp;$\varepsilon$, we have $L'_0L'_1 \supseteq L'_0\cup L'_1 = \{0,1\}^*\!$.
The class of NP-complete languages is not closed under Kleene star.  For any NP-complete language&nbsp;$L$, $L\cup \{0,1\}$ is NP-complete but $\big(L\cup \{0,1\}\big)^* = \{0,1\}^*\!$.
If the  class of NP-complete problems is closed under complementation, then NP&nbsp;=&nbsp;coNP. Whether this is true or not is one of the major open problems in complexity theory.