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**Dinkydoe** I can better paint the picture by making it more concrete: Let $\displaystyle A = \left\{a_1,a_2,\cdots \right\}$. Then $\displaystyle A\cup \left\{\emptyset\right\}=\left\{\emptyset, a_1,a_2,\cdots \right\} $

Wrong!

$\displaystyle \emptyset \in \left\{\emptyset\right\}$. The empty set is an element of this set. You can write the emptyset as: $\displaystyle \emptyset = \left\{\right\}$. It's a container without elements. But a container without elements, can still be an element of another container! Namely: $\displaystyle \left\{\emptyset\right\} = \left\{\left\{\right\}\right\} $

And how do you conclude $\displaystyle \emptyset \not\in A$? The empty-set may, or may not be in $\displaystyle A$. It depends on how we define $\displaystyle A$ to be.

$\displaystyle \emptyset, \left\{\emptyset\right\}, \left\{\left\{\emptyset\right\}\right\}$ all may, or may not be contained in $\displaystyle A$. We didn't specify $\displaystyle A$, so we don't know.