# set theory - trivial question

If I have a set $\Omega = \{ \{1 \},\{2\} \}$ is it true that $\{1,2\}$ belongs to $\Omega$? I guess it is true because $\{a \} \cup \{b \} = \{a,b\}$. Then what is confusing me is if I have a power set $P(\Omega ) = \{ \emptyset ,\{ 1\} ,\{ 2\} ,\{ 1,2\} \}$ why do I need to have explicitly listed $\{1,2\}$ as a particular element of $\Omega$ when it is already ensured by the presence of the sets $\{ 1\} ,\{ 2\}$.

• But $\{a,b\} \ne \{\{a\},\{b\}\}$. – peterwhy Dec 13 '14 at 9:23
• Also you made a mistake $\mathcal P(\Omega ) \neq \{ \emptyset ,\{ 1\} ,\{ 2\} ,\{ 1,2\} \},$ the correct expression is $\mathcal P(\Omega)=\big\{\emptyset,\{\{1\}\},\{\{2\}\},\{\{1\},\{2\}\}\big\}.$ – Hakim Dec 13 '14 at 9:30
• If I have $\{ 1,2\} = \{ 1\} \cup \{ 2\}$ can't I use a similar principle to $\{ \{ 1\} ,\{ 2\} \}$? Then I should get $\{ \{ 1\} ,\{ 2\} \} = \{ \{ 1\} \cup \{ 2\} \} = \{ \{ 1,2\} \}$ – Fragile Dec 13 '14 at 9:32
• @Fragile But $\{1\}\cup\{2\}=\{1,2\}\neq\{1\},\{2\}.$ – Hakim Dec 13 '14 at 9:34
• I think the mistake I made was the following: $\{ \{ 1\} ,\{ 2\} \} = \{ \{ 1\} \} \cup \{ \{ 2\} \}$ but $\{ \{ 1\} ,\{ 2\} \} \ne \{ \{ 1\} \cup \{ 2\} \}$, which is what I assumed above. Am I correct? – Fragile Dec 13 '14 at 9:51

If $\Omega=\{\{1\},\{2\}\}$, then $\Omega$ contains only two elements: $\{1\}$ and $\{2\}$. Since $\{1\}\ne\{1,2\}$ and $\{2\}\ne\{1,2\}$, then $\{1,2\}$ is not an element of $\Omega$.

Each "subset" is a element of $\Omega$, i.e, $\Omega$ is a set of sets, then

$P(\Omega) = \{\emptyset,\{\{1\}\} ,\{\{2\}\} , \Omega \}$

• Yes ! thanks you ! – user6565190 Dec 13 '14 at 9:30

Belonging relation is not transtive, that is, you can have that $x\in X$ and $X\in\mathcal X$ but $x\notin\mathcal X$.

For example, $1\notin\{\{1\}\}$

• This is not quite the issue here, but it was the issue on a question yesterday. – Asaf Karagila Dec 13 '14 at 9:32