Define the $\mathcal P \left(\cup\{\{\emptyset\}\}\right)$ set. What is the $\mathcal P \left(\cup\{\{\emptyset\}\}\right)$ set?
 A: As usually, when these problems my advice is to use placeholders for the actual sets, and analyze this step by step.
Let's set $a=\{\varnothing\}$ and $b=\{\{\varnothing\}\}$ and $c=\bigcup b$. We are asked, if so, to find out $\mathcal P(c)$. What is this set? It's the set $\{x\mid x\subseteq c\}$.
So first order of business is to understand what is $c$. Recall that $\bigcup X=\{y\mid\exists x\in X.y\in x\}$. So $c=\bigcup b=\{y\mid\exists x\in b.y\in x\}$. So what are the elements of $b$? Just $a$ itself. So what is $c$? It's the set of all the elements of $a$. This means exactly that $c=a$.
So we are asked to find out what is $\mathcal P(a)$. And I will leave you with the task of finding out the contents of this set in details.
A: See Arbitrary union :


union of an arbitrary collection of sets, sometimes called an infinitary union. If $M$ is a set whose elements are themselves sets, then $x$ is an element of the union of $M$ if and only if there is at least one element $A$ of $M$ such that $x$ is an element of $A$. In symbols:
$x \in \cup M \leftrightarrow \exists A \in M, x \in A$.


In our example, $M = \{\{\emptyset\}\}$ and its only element $A$ is $\{\emptyset\}$.
Which are the elements of $\{\emptyset\}$ ? Only $\emptyset$.
Thus, $x \in \cup M \leftrightarrow x = \emptyset$, i.e. $\cup M = \{\emptyset\}$.
In conclusion : $\mathcal P \left(\cup\{\{\emptyset\}\}\right) = \mathcal P(\{\emptyset\})$ .
A: Why not just calculate which elements $\;V\;$ are in this set, by expanding the definitions, as follows:
\begin{align}
& V \in \mathcal P \left(\bigcup \left\{\left\{\emptyset\right\}\right\}\right) \\
\equiv & \qquad \text{"definition of $\;\mathcal P\;$"} \\
& V \subseteq \bigcup \left\{\left\{\emptyset\right\}\right\} \\
\equiv & \qquad \text{"definition of $\;\subseteq\;$"} \\
& \langle \forall x :: x \in V \;\Rightarrow\; x \in \bigcup \left\{\left\{\emptyset\right\}\right\} \rangle \\
\equiv & \qquad \text{"definition of $\;\bigcup\;$"} \\
& \langle \forall x :: x \in V \;\Rightarrow\; \langle \exists A : A \in \left\{\left\{\emptyset\right\}\right\} : x \in A \rangle \rangle \\
\equiv & \qquad \text{"definition of $\;\left\{\dots\right\}\;$"} \\
& \langle \forall x :: x \in V \;\Rightarrow\; \langle \exists A : A = \left\{\emptyset\right\} : x \in A \rangle \rangle \\
\equiv & \qquad \text{"logic: simplify using one-point rule"} \\
& \langle \forall x :: x \in V \;\Rightarrow\; x \in \left\{\emptyset\right\} \rangle \\
\equiv & \qquad \text{"..."} \\
\end{align}
Now reintroduce to $\;\subseteq\;$ and see which $\;V\;$ you find.
