# Relations notation

say, you have $R^* :== \bigcup_{k = 0}^{\infty}R^k$, where $R$ relation and $R^n =R^{n-1}R$.

Then $(R^*)^* = (\bigcup_{k = 0}^{\infty}R^k)^* =$ ??

I am not sure about how to open the second $*$

By definition, $$(R^*)^* = \bigcup_{k = 0}^{\infty} (R^*)^k = \bigcup_{k = 0}^{\infty} ( \bigcup_{l=0}^{\infty} R^l ){}^k$$
Now (can you see and show this?), $$( \bigcup_{l=0}^{\infty} R^l )^k = \bigcup_{l=0}^{\infty} R^{lk}$$ so$$(R^*)^* = \bigcup_{k = 0}^{\infty} (R^*)^k = \bigcup_{k = 0}^{\infty} \bigcup_{l=0}^{\infty} R^{lk} = \bigcup_{n = 0}^{\infty} R^{n} = R^*$$ where the reduction from two unions to one union is valid since $lk$ runs over all natural numbers.
• And, what that could mean? The union of all $R^l$-s is a one set. What is the union over all $k$-s over one set? – Kirill Nov 5 '17 at 19:52