Show that $a\in X\Longrightarrow\mathcal{P}(a)\in\mathcal{P}(\mathcal{P}(\bigcup X))$ Do not understand the solution I have been given,
$\mathcal{P}(a)\in\mathcal{P}(\mathcal{P}(\bigcup X))\Longleftrightarrow\mathcal{P}(a)\subseteq\mathcal{P}(\bigcup X))$ [simplifying RHS of equation - understand]
Let $t\subseteq a$
then,
$a\in X\Longrightarrow [t\subseteq a\Longrightarrow t\subseteq\bigcup X\Longleftrightarrow t\in\mathcal{P}(\bigcup X)]$ [understand this]
also,
$t\subseteq a\Longleftrightarrow t\in\mathcal{P}(a)$ [understand this]
therefore
$\mathcal{P}(a)\subseteq\mathcal{P}(\bigcup X)$ [dont understand this]
Surely we have shown $\mathcal{P}(a)=\mathcal{P}(\bigcup X)$, why is one a subset of the other, and not the other way around?
 A: I don't buy the line
$$
a \in X \Rightarrow \left[t \subset a \Leftrightarrow t \subset \bigcup X \Leftrightarrow t \in \mathcal{P}\left(\bigcup X \right)\right].
$$
That 
$$
t \subset a \Rightarrow t \subset \bigcup X
$$ 
is clear, but 
$$
t \subset \bigcup X \Rightarrow t \subset a
$$ 
doesn't hold in general. For example, $X=\{\{0\},\{1\}\}$, $a=\{0\}$ and $t=\{1\}$.
A: The reason for you not understanding is exactly the mistake that the first answer pointed out.

$
\newcommand{\ref}[1]{\text{(#1)}}
$In your original question you had an incorrect "proof" of
$$
\tag 1
\langle \forall t :: t \subseteq a \iff t \in \mathcal P(\bigcup X) \rangle
$$
and stated (correctly) that
$$
\tag 2
\langle \forall t :: t \subseteq a \iff t \in \mathcal P(a) \rangle
$$
Combining those gives
$$
\tag 3
\langle \forall t :: t \in \mathcal P(a) \iff t \in \mathcal P(\bigcup X) \rangle
$$ which (by set extensionality) is the same as $\;\mathcal P(a) = P(\bigcup X)\;$, as you expected.

After the answer by Tom, you corrected $\ref 1$ to
$$
\tag{1'}
\langle \forall t :: t \subseteq a \implies t \in \mathcal P(\bigcup X) \rangle
$$
From $\ref{1'}$ and $\ref 2$ we cannot conclude $\ref 3$ anymore: now we get
$$
\tag{3'}
\langle \forall t :: t \in \mathcal P(a) \implies t \in \mathcal P(\bigcup X) \rangle
$$ which (by the definition of $\;\subseteq\;$) is the same as $\;\mathcal P(a) \subseteq P(\bigcup X)\;$, as the given solution says.
A: Suppose $a\in X$. We need to show $\mathcal P (a)\subseteq \mathcal P (\bigcup X)$. Well, let $b\subseteq a$. Note that $c\in b$ implies $c\in\bigcup X$. So $b\subseteq \bigcup X$.
