# real analysis – Help with Baby Rudin Definition 11.4   Hey all, I got stuck on the proof of equation $$(12)$$. My consideration is as follows:
Assume that $$Ain {scr E}$$ and $$Bin {scr E}$$. Then $$A=bigcup_{s=1}^n I_s$$, $$B=bigcup_{k=1}^m J_k$$, where $$I_s$$ and $$J_k$$ are intervals in $$R^p$$ for $$s=1,2,cdots,n$$ and $$k=1,2,cdots,m$$.
Since $$Abigcup B=I_1bigcupcdotsbigcup I_nbigcup J_1bigcupcdotsbigcup J_m$$, $$Abigcup Bin {scr E}$$. It is left to show that $$A-Bin {scr E}$$.
Note that
begin{aligned} A-B&=Abigcap B^c\ &=Abigcapleft(bigcup_{k=1}^m J_kright)^c\ &=Abigcapleft(bigcap_{k=1}^m J_k^cright) (by De Morgan’s Law)\ &=bigcap_{k=1}^mleft(Abigcap J_k^cright)\ &=bigcap_{k=1}^m left( bigcup_{s=1}^n left(I_sbigcap J_k^c right)right) end{aligned}
It can be shown that $$bigcap_{k=1}^m left( bigcup_{s=1}^n left(I_sbigcap J_k^c right)right)=bigcup_{s=1}^n left( bigcap_{k=1}^m left(I_sbigcap J_k^c right)right)$$, so we only need to show $$bigcap_{k=1}^m left(I_sbigcap J_k^c right)$$ is an interval.
Now we prove the intersection of any two intervals(namely, $$A$$ and $$B$$) is also an interval:
If $$Abigcap B=0$$, then it is an interval by definition.
If $$Abigcap Bneq 0$$, then the points $$mathbf x=(x_1,x_2,cdots,x_p)$$ in $$Abigcap B$$ has the form
$$c_ileq x_i leq d_iquad (i=1,2,cdots,p),$$
so that $$Abigcap B$$ is an interval.
If we can show that the complement of an interval is also an interval, then the theorem completes. This is where I got stuck. Furthermore, I have no idea about how to show $${scr E}$$ is not a $$sigma$$-ring.
I would appreciate if you could explain in details. 