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If $A_\alpha$ are subsets of a set $S$ then

$\bigcup_{\alpha \in I}A_\alpha$ = "all $x \in S$ so that $x$ is in at least one $A_\alpha$"

$\bigcap_{\alpha \in I} A_\alpha$ = "all $x \in S$ so that $x$ is in all $A_\alpha$"

It is the convention that $\bigcup_{\alpha \in \emptyset}A_\alpha = \emptyset$ and $\bigcap_{\alpha \in \emptyset} A_\alpha = S$.

But if $x$ is in $\bigcap_{\alpha \in \emptyset} A_\alpha = S$ then $x$ is in all $A_\alpha$ with $\alpha \in \emptyset$ and therefore $x$ is certainly in at least one $A_\alpha$ with $\alpha \in \emptyset$. But then $x \in \bigcup_{\alpha \in I}A_\alpha$.

Can someone help me and tell me what is wrong with this? Thank you.

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This might help. – Gautam Shenoy Apr 23 '13 at 9:13
Element in the empty set is the problem. – simplicity Apr 23 '13 at 9:14
up vote 14 down vote accepted

Some texts consider it a convention, but it is in fact a computation!

For the fixed set $S$, we are looking at the set $\mathcal P(S)$, the set of all subsets of $S$. With the operations of intersection and unions (of arbitrary many subsets) the set $\mathcal P(S)$ is what is known as a complete lattice (don't worry if you don't know what that means).

First, one can argue intuitively: the more subsets of $S$ you intersect, the smaller the intersection is. Or, the fewer subsets you intersect, the larger the intersection is. So, the intersection of no subsets at all, the least amount of sets you can intersect, should be the largest subset possible. Namely, $\bigcap_{i\in \emptyset}A_i=S$. Similarly, the fewer subsets you take the union of, the smaller the union. So, the union of no subsets at all should be the smallest set possible. Namely, $\bigcup _{i\in \emptyset }A_i=\emptyset$.

Now, to make things more formal, lets define the intersection and union in $\mathcal P(S)$. The definition will be equivalent to the set-theoretic definitions but will only make use of the partial order relation of inclusion. Given a collection $\{A_i\}_{i\in I}$ of subsets of $S$, their intersection is the largest subset of $S$ that is contained in each $A_i$ (notice that this is saying that the intersection is a greatest lower bound). Similarly, the union of the family of subsets is the smallest subset of $S$ that contains each $A_i$ (notice that this says that the union is a least upper bound). Incidentally, this point of view very clearly points to a duality between union and intersection.

So now, the intersection of no subsets is the largest subset of $S$ that is contained in each one of the given subsets. There are no given subsets at all, so (vacuously) any subset $B\subseteq S$ is contained in each of the non-existent $A_i$. The largest of those is $S$, proving that $\bigcap_{i\in \emptyset}A_i=S$. Similarly, the union of no subsets is the smallest subset of S that contains each of the given subsets. No subsets are given, so any subsets $B\subseteq S$ contains each of the non-existent $A_i$. The smallest of these is $\emptyset $, thus proving that $\bigcup_{i\in \emptyset}A_i=\emptyset$.

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Please note that the fact that

$$\bigcap_{\alpha \in \emptyset} A_\alpha = S$$

is not a convention - it follows from the definition

$$\bigcap_{\alpha \in I} A_\alpha = \{ x \in S : \text{$x \in A_\alpha$, for all $\alpha \in I$}\}.$$

This is best understood by asking when is it that for a given $x \in S$ we have $x \notin \bigcap_{\alpha \in \emptyset} A_\alpha$. This can only happen if there exists $\alpha \in \emptyset$ such that $x \notin A_{\alpha}$. Since $\emptyset$ has no elements, there cannot be such an $\alpha$.

PS One can avoid introducing a (somewhat arbitrary) index set in an intersection, say. Instead, fix a set $S$, take a subset $\mathfrak{S}$ of $\mathcal{P}(S)$, and define $$ \bigcap \mathfrak{S} = \{ x \in S : \text{$x \in A$ for all $A \in \mathfrak{S}$} \}. $$ Then the argument above becomes $$ \bigcap \emptyset = \{ x \in S : \text{$x \in A$ for all $A \in \emptyset$} \} = \{ x \in S : \ \} = S,$$ since there are no $A$ to consider here.

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It should be pointed out that this intersection is $S$ only if it is given that the computation is performed in $\mathcal P(S)$. Otherwise, if the computation is performed just in the ambient universe of sets, then the intersection does not exist (since it's trying to be a set that contains every other set). In contrast, the empty union is the empty set no matter where it is computed. – Ittay Weiss Apr 23 '13 at 9:47
@DanDouglas In the case of the union of an empty collection, you are considering $U = \{ x \in S : \text{there exists $A \in \emptyset$ such that $x \in A$} \}$. Now no $x \in S$ can satisfy the condition, as there is no $A \in \emptyset$. In other words, "there exists $A \in \emptyset$ such that ..." is always false. – Andreas Caranti Dec 13 '13 at 8:38
@DanDouglas, Ittay Weiss had already clarified this point in a comment to another answer. – Andreas Caranti Dec 13 '13 at 8:44
@DanDouglas, just consider its negation "there exists $x \in \emptyset \dots$" – Andreas Caranti Dec 20 '13 at 21:38

Those conventions comes from the fact that $(\mathcal P (S), \cup, \emptyset)$ and $(\mathcal P (S), \cap, S)$ are monoids.

Hence, those conventions are just the usual $\sum_{k \in \emptyset} k = 0$ or $\prod_{k \in \emptyset} k = 1$ that you certainly know for $(\mathbb N, +)$ or $(\mathbb Z, \times)$ for example.

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This is like saying "it's so because it is so". In fact, it is not a convention at all. It can be proven so by following the definition. In a monoid though, the unit is (by definition if you like) the result of applying the operation to no elements. But for $\mathcal P(S)$ it follows from a deeper reason than just saying it's so. – Ittay Weiss Apr 23 '13 at 9:34
@IttayWeiss The question was not "Is it a convention or not ?" but "Can anyone enlighten me about those conventions/properties ?". I was just pointing out that the OP was actually familiar with those but did not recognize it at first sight. Plus, I could argue that your proof can actually show that $\bigcap_\emptyset A_i = \emptyset$ : indeed, taking $B$ a subset of $S$, $\forall A_i \in \emptyset, A_i \not\subseteq B$ is as true as $\forall A_i \in \emptyset, A_i \subseteq B$, making $\emptyset$ the intersection over the emptyset. – Pece Apr 23 '13 at 19:16
I see your your points with the monoid analogy. If you'll think about it though, you'll see that your criticism of my proof is false (or are you claiming that you just proved a contradiction?) – Ittay Weiss Apr 23 '13 at 19:45
Sorry, I mixed up a little. What I was saying is : as $A_i \in \emptyset$ is universally false, your claim $\forall A_i, A_i \in \emptyset \implies B \subseteq A_i$ (and so $S$, the greatest of all, is the intersection) is a valid as my claim $\exists A_i, A_i \in \emptyset \implies B \not\subseteq A_i$ (and so no subset of $S$ is a lower bound of the $A_i$ except $\emptyset$). So in my opinion, you're making a convention on the complete lattices (the meet of nothing is the top) while I was making conventions about monoid. – Pece Apr 24 '13 at 8:34
I think you are still confused. I did not make use of any convention. I followed the definition. – Ittay Weiss Apr 24 '13 at 8:56

Some sybolic manipulation can help, to see what is NOT in those sets,

$ x \in \ \bigcap_{\alpha \in I} A_\alpha \ \Longleftrightarrow \forall\alpha\in I,\ x\in A_\alpha \\ x \in \ \bigcap_{\alpha \in I} A_\alpha \ \Longleftrightarrow \forall\alpha, \ \alpha\in I\rightarrow\ x\in A_\alpha \\ x \notin \ \bigcap_{\alpha \in I} A_\alpha \ \Longleftrightarrow \exists\alpha: \ \alpha\in I\ \wedge\ x\notin A_\alpha \\ x \notin \ \bigcap_{\alpha \in \emptyset} A_\alpha \ \Longleftrightarrow \exists\alpha: \ \alpha\in \emptyset\ \wedge\ x\notin A_\alpha $

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That is not a bad answer, but the meaning is not immediately clear. Could you expand on the meaning of the last line and how it ties to the question? – Rory Daulton Dec 13 '15 at 23:39

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