Does $\bigcup\emptyset$ equal $\bigcup\{\emptyset\}$? For any collection of sets $A = \{A_i\ : i \in I \}$, define
$$\bigcup A = \bigcup_{i \in I} A_i$$
Question: Is the following true?
$$\bigcup \emptyset = \bigcup \{\emptyset \}$$
The right-hand side, using the definition, is simply the empty set.  However, I'm not sure how I'm supposed to interpret the left - hand side.  I'm guessing the fact that the left-hand side is the empty set follows "vacuously", but I'm not too certain.
Can somebody help me understand?
 A: The union on the RHS is the empty set: if $x$ is in the union, then $x\in \emptyset$, a contradiction, thus there are no elements in the union.
The union on the LHS, as a union over an empty indexed collection of sets $\{A_i\}_{i\in I}$ with $I=\emptyset$ is also the empty set: if $x$ is in the union, then there must be some index $i\in I$ with $x\in A_i$. Since no such $i$ exists, there are no elements in the union. 
A: Recall that (by definition, see, for example, Wikipedia)
$$x\in\bigcup \mathcal S \Leftrightarrow (\exists S\in \mathcal S)x\in S$$
So $\bigcup\emptyset=\emptyset$, since for no element $x$ there exists $S\in\emptyset$ fulfilling $x\in S$. (Simply because there is no $S$ in $\emptyset$.)
A: The union $\bigcup\varnothing$ is empty vacuously (the union of no sets).
The union $\bigcup\{\varnothing\}$ is empty because $\bigcup\{x\}=x$, and in this case $x$ is the empty set.
So yes, they are equal.
Note that this is not a trait unique to the empty set. Take $\omega$ for example, the first infinite ordinal. $\omega$ has the property $\omega=\bigcup\omega$, so we also have $\bigcup\omega=\bigcup\{\omega\}$. The same is true for any limit ordinal.
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$To expand on the other answers, here is a slightly more detailed proof that both sets are empty, but for different reasons.
For the first set, we calculate its members $\;x\;$ as follows:
$$\calc
x \in \bigcup \emptyset
\calcop{\equiv}{definition of $\;\bigcup\;$}
\langle \exists V : V \in \emptyset : x \in V \rangle
\calcop{\equiv}{definition of $\;\emptyset\;$}
\langle \exists V : \text{false} : x \in V \rangle
\calcop{\equiv}{logic: simplify}
\text{false}
\endcalc$$
For the second set, we similarly calculate for all $\;x\;$:
$$\calc
x \in \bigcup \{\emptyset\}
\calcop{\equiv}{definition of $\;\bigcup\;$}
\langle \exists V : V \in \{\emptyset\} : x \in V \rangle
\calcop{\equiv}{definition of $\;\{\ldots\}\;$}
\langle \exists V : V = \emptyset : x \in V \rangle
\calcop{\equiv}{logic: one-point rule}
x \in \emptyset
\calcop{\equiv}{definition of $\;\emptyset\;$}
\text{false}
\endcalc$$
Therefore both sets are empty, and therefore they are equal.
