$\{a^n b^n\} \cap \{a^*b^*\}$ regular or not?

I have been asked a very interesting question in preparation for my exam coming up. The question is as follows:

Is $L=\{a^n b^n : n \geq 0\} \cap \{a^*b^*\}$ regular or not?

Assume $L$ is regular. Then, $L^c$ should be regular as well. Thus, $L^c = \{a^n b^n : n \lt 0\} = \{\}$, so if I compliment the compliment, I should get $L = U$ (the universal set). So, then $L=\{a^n b^n : n \geq 0\} \cap \{a^*b^*\} \neq U$, because $L$ does not contain words of the form $baba$, or $baaaababa$, etc. Therefore the language is not regular.

Is this an acceptable answer for the upcoming exam? Does this persuade anyone?

• INTERSECTION of a * b * not ab – bmc Oct 5 '16 at 17:30
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The complement of $\{a^nb^n \mid n \geq 0\}$ is not $\{a^nb^n \mid n < 0\}$. And the complement of the complement should always equal the original set. The rest of your argument falls apart here.
As an example, abb is a string that is not $a^n b^n$, thus it should be in the complement of L.
To hint at an answer, is there any string $a^nb^n$ that is not $a^*b^*$? What does this say about the intersection of the two?
• Correct, the intersection is $a^nb^n$. The intersection with $a^*b^*$ is just to throw you a bit of a curveball, and you can use the standard pumping lemma proof after finding out the intersection. – orlp Oct 5 '16 at 18:08