Relating the union of the power set of a set to the set itself I'm pretty sure I've asked way too many questions on sets already(still, a big thanks to everyone who helped) and even though I'm starting to see what's going on there are still some issues, for example this question:
Prove that $\bigcup P(X)=X$, where $P(X)$ is the power set for the set $X$
So, what I can do is the following:
$a \in \bigcup P(X) \implies \exists b \subset X$ s.t. $a\in b \implies a \in X$ and for the other direction I can take $b=X$.
However, I'm not sure whether this is correct, and in particular, the bit with $b \subset X, a\in b \implies a \in X$
If I take $X=\{\{1\},\{2\}\}$, then $P(X)=\{\emptyset,\{1\},\{2\},\{\{1\},\{2\}\}\}$, hence $\bigcup P(X)=\{1,2,\{1\},\{2\}\}$ which is not equal to $X$ by ZF$1$
Where's the problem?
Thanks
 A: Let's approach proving the equality
$$
\bigcup \mathcal P(X) = X
$$
using two inclusions:
"$\subseteq$":
Let $x \in \bigcup \mathcal P(X)$.
Then there exists a subset $A \in \mathcal P(X)$ of $X$ with $x \in A$. Now:
 $$A \subseteq X \text{ and } x \in A \implies x \in X$$
"$\supseteq$:" Let $x \in X$. Then $\{x\} \subseteq X$ is a subset of $X$. Therefore $\{x\} \in \mathcal P(X)$. Now:
$$
x \in \{x\} \text{ and } \{x\}\in \mathcal P(X) \implies x \in \bigcup \mathcal P(X)
$$
Your example now looks like this:
$$
X = \{\{1\}, \{2\}\} = \emptyset \,\cup\, \{\{1\}\} \,\cup\, \{\{2\}\} \,\cup\, \{\{1\},\{2\}\} = \bigcup \mathcal P(X)
$$
A: Here is a good way to see through the confusion, which generally stems from not seeing the set from the braces.
Replace $\{1\}$ by $a$ and $\{2\}$ by $b$. Now $X=\{a,b\}$ and $\mathcal P(X)=\{\varnothing,\{a\},\{b\},\{a,b\}\}$.
What you wrote in your post is actually $\{\varnothing,a,b,\{a,b\}\}$, and since $a\neq\{a\}$ and $b\neq\{b\}$ (and also $a\neq\{b\}$ and $b\neq\{a\}$), this is not the same as the power set of $X$.
To see that the result holds, not that if $A\in\mathcal P(X)$, then $A\subseteq X$. From this the result follows.
