# Complement of $0^n1^n | n \in \mathbb{N}$

I know why A is irregular by Closure properties of irregular language. I also know the complement of $$\{ 0^n 1^n | n \in \mathbb{N}\}$$ is $$A = \{ 0^i 1^j| i \neq j\} \cup (0 \cup1)^*(1)(0 \cup1)^*0(0 \cup1)^*$$, but I don't see how $$A^* = \sum^*$$. Any help would be appreciated!

• You did a fine job presenting "your own formulas" using MathJax: please make the problem statement (?) a block quote, consider presenting it using MathJax (searchable) instead of hyperlinking a pixel raster. Jun 11 '20 at 9:41

Since $$0 \in A$$ and $$1 \in A$$, any word $$w \in \Sigma^*$$ can be written has a concatenation of $$n=|w|$$ words $$w_1, \dots, w_n$$ such that each $$w_i \in A$$ (showing that $$w \in A^*$$). In order to do so it suffices to pick $$w_i$$ as the $$i$$-th character of $$w$$, i.e., $$w_i$$ is either $$0$$ or $$1$$.