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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 $\{ 1\} ,\{ 2\}$.

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

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

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

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  • $\begingroup$ Yes ! thanks you ! $\endgroup$ – user6565190 Dec 13 '14 at 9:30
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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, $1\notin\{\{1\}\}$

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

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