# Does $\Sigma^* \cdot a^nb^n=\Sigma^*$

Is it true to say that $$\Sigma^* \cdot$$ {$$a^nb^n: n>=0$$} = $$\Sigma^*$$

Becuase if we take $$\Sigma^*$$ and concatenate it to {$$a^nb^n: n>=0$$} we don't get any "new" words than those we had in $$\Sigma^*$$ in the first place.

• yes, it is as $n \geq 0$. – OmG Dec 15 '18 at 15:29

Yes if $$\Sigma \supseteq \{a, b\}$$. To show the equality, let's show one is a subset of the other, and vice versa.
• $$\Sigma^\ast \cdot \left\{ a^n b^n \middle| n \geq 0 \right\} \subseteq \Sigma^\ast$$ holds, because for any word $$w \in \Sigma^\ast \cdot \left\{ a^n b^n \middle| n \geq 0 \right\}$$, we know $$w = u a^n b^n$$ for $$u \in \Sigma^\ast$$, therefore $$w \in \Sigma^\ast$$ since $$\Sigma \supseteq \{a, b\}$$.
• $$\Sigma^\ast \cdot \left\{ a^n b^n \middle| n \geq 0 \right\} \supseteq \Sigma^\ast$$ holds, because
$$\Sigma^\ast \cdot \left\{ a^n b^n \middle| n \geq 0 \right\} \supseteq \Sigma^\ast \cdot \{ a^0 b^0 \} = \Sigma^\ast \cdot \{ \varepsilon \} = \Sigma^\ast$$
where $$\varepsilon$$ is the empty word.