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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?

Asaf Karagila
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Bart
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    Related: http://math.stackexchange.com/questions/370188/empty-intersection-and-empty-union – Jonas Meyer May 07 '13 at 04:05
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    I think the RHS is ${\emptyset}$, not $\emptyset$; while the LHS is just $\emptyset$. Remember, $\cup A_i = {x : x\in A_i$ for some $i.}$, hence $\emptyset$ is an element of the RHS. – tghyde May 07 '13 at 04:07
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    @thyde641 The symbol on RHS denotes taking the union of all elements in the set, so it would be just the empty set, not the set containing the empty set. – Bart May 07 '13 at 04:27
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    I have changed the title, so that it better describes the question. If needed, edit/improve it again. It is useful to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Martin Sleziak May 07 '13 at 06:59

4 Answers4

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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.

Asaf Karagila
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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.

Ittay Weiss
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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 $\emptyset$.)

Martin Sleziak
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$ \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.

MarnixKlooster ReinstateMonica
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