sigma field on empty set, and on set containing empty set. Given $\Omega=\{1,2\}$, and $\mathscr E_1=\emptyset$,$\mathscr E_2=\{\emptyset\}$ Then i want to generate the smallest sigma field for each given case. The answer is:
$$
\mathscr F_1 = \{\emptyset,\emptyset^c=\Omega\}
$$
$$
\mathscr F_2 = \{\emptyset,\Omega\}
$$
That is, $\mathscr F_1 = \mathscr F_2$. However, I find this to be counterintuitive, as $\emptyset \ne \{\emptyset\}$, so shouldn't it be:
$$
\mathscr F_2^s = \{\emptyset,\Omega,\{\emptyset\},\{\Omega\}\}
$$
Although I know this doesn't make sense either as $\mathscr F_2^s$ is not a subset of the power set of $\Omega$....

The OP, as far as I can understand, insists that we ought to differentiate between the two $\sigma$-algebras. At the same time the OP admits that he couldn'n describe the right solution. 
Here is my answer:
(1) By definition, a $\sigma$-algebra is a nonempty class. That is $$\sigma(\emptyset)=\sigma(\mathscr E_1)\not=\emptyset.$$
(2) By definition, the $\sigma$-algebra generated by a class is the intersection of all $\sigma$-algebras containing that class. As a result
$$\sigma(\emptyset)=\sigma(\mathscr E_1)=\{\emptyset,\Omega\}.$$
and for the same reason$$\sigma(\{\emptyset\})=\sigma(\mathscr E_2)=\{\emptyset,\Omega\}.$$
This is so, because $$\emptyset \subset \text{of any set},$$
at the same time $$\{\emptyset\} \subset \text{of any }\sigma\text{ algebra}.$$
