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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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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 union (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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    $\begingroup$ Why not just use the definition: $x\in\bigcap_{i\in\emptyset}A_i\Leftrightarrow\forall_i(i\in\emptyset\Rightarrow x\in A_i)$? From this we can conclude that every $x\in S$ satisfies the right-hand condition, and thus $\bigcap_{i\in\emptyset}A_i=S$. $\endgroup$ – Mad Hatter Mar 21 '17 at 20:58
  • $\begingroup$ I proved it using Mad Hatter's method, but I appreciate the intuition provided above. $\endgroup$ – EternusVia Jan 20 '18 at 2:33
  • $\begingroup$ @MadHatter since the reasoning above holds in any complete lattice, showing what the empty join and empty meet must be. No element-wise argument can be given in a general lattice, since its elements may not be sets, nor is the ordering necessarily given by set inclusion. $\endgroup$ – Ittay Weiss Feb 20 '18 at 9:25
  • $\begingroup$ If you want it for any complete lattice it is enough to notice that the unity of the lattice L is tha largest lower bound of the empty subset of L. $\endgroup$ – Mad Hatter Feb 22 '18 at 13:32
  • $\begingroup$ @MadHatter that is exactly the point. $\endgroup$ – Ittay Weiss Feb 22 '18 at 14:59
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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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    $\begingroup$ 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. $\endgroup$ – Ittay Weiss Apr 23 '13 at 9:47
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    $\begingroup$ @Anna, see en.wikipedia.org/wiki/Universal_set and en.wikipedia.org/wiki/Russell%27s_paradox $\endgroup$ – Andreas Caranti Apr 23 '13 at 10:07
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    $\begingroup$ @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. $\endgroup$ – Andreas Caranti Dec 13 '13 at 8:38
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    $\begingroup$ @DanDouglas, Ittay Weiss had already clarified this point in a comment to another answer. $\endgroup$ – Andreas Caranti Dec 13 '13 at 8:44
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    $\begingroup$ @DanDouglas, just consider its negation "there exists $x \in \emptyset \dots$" $\endgroup$ – Andreas Caranti Dec 20 '13 at 21:38
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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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  • $\begingroup$ 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. $\endgroup$ – Ittay Weiss Apr 23 '13 at 9:34
  • $\begingroup$ @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. $\endgroup$ – Pece Apr 23 '13 at 19:16
  • $\begingroup$ 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?) $\endgroup$ – Ittay Weiss Apr 23 '13 at 19:45
  • $\begingroup$ 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. $\endgroup$ – Pece Apr 24 '13 at 8:34
  • $\begingroup$ I think you are still confused. I did not make use of any convention. I followed the definition. $\endgroup$ – Ittay Weiss Apr 24 '13 at 8:56
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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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  • $\begingroup$ 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? $\endgroup$ – Rory Daulton Dec 13 '15 at 23:39
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But if "$\bigcap_{\alpha \in \emptyset} A_\alpha = S$" then "$x$ is in all $A_\alpha$ with $\alpha \in \emptyset$ and therefore $x$ is certainly in at least one $A_\alpha$ with $\alpha \in \emptyset$", meaning then $x \in \bigcup_{\alpha \in I}A_\alpha=S$.

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

This is a bit late, but recall that when dealing with an empty domain, universality does not imply existence.

$$\forall \alpha\in\emptyset ~(P(\alpha)) ~\nvdash~ \exists \alpha\in\emptyset~(P(\alpha))$$

And the definition of the empty intersection is: $~\bigcap_{\alpha\in\emptyset} A_\alpha~{=\{x\in S:\forall \alpha\in\emptyset~(x\in A_\alpha)\}\\=\{x\in S:\top\}\\=S}$

While the definition of the empty union is: $\qquad\bigcup_{\alpha\in\emptyset} A_\alpha~{=\{x\in S:\exists \alpha\in\emptyset~(x\in A_\alpha)\}\\=\{x\in S:\bot\}\\=\emptyset}$

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Note that actually, $\cap_{i\in \emptyset} A_i = S$ is not a convention, it's actually just wrong.

The correct statement would be $\cap_{i\in \emptyset} A_i = \emptyset$, or $\bigwedge_{i\in \emptyset} A_i = S$, where $\bigwedge$ denotes the greatest lower bound in the complete lattice $(\mathcal{P}(S), \subset)$.

Why are these true ? Well first note that $\cap_{i\in \emptyset} A_i = S$ simply cannot be true. One the LHS we have an intersection of sets with no mention of $S$ (the $A_i$'s are subsets of $S$, but they're also subsets of $S\cup \{S\}$ or god knows what), and on the RHS we have a set $S$, that happens to contain the $A_i$'s. So thats just wrong because then, $\cap_{i\in \emptyset} A_i = S\cup\{S\}$ would also be true (the $A_i$s are subsets of $S\cup\{S\}$, after all), which is absurd.

So $\bigcap_{i\in I}A_i$ is actually an abbreviation for $\bigcap \{A_i \mid i\in I\}$ which is by definition $\{x\in \displaystyle\bigcup_{i\in I} A_i \mid \forall i\in I, x\in A_i\}$, where $\displaystyle\bigcup_{i\in I}A_i$ is an abbreviation for $\bigcup\{A_i\mid i\in I\}$ which is $\{x\mid \exists i\in I, x\in A_i\}$ (its existence is provided by the axiom of reunion). But then if $I=\emptyset$, this last thing is empty, and so are all its subsets, in particular $\cap_{i\in \emptyset} A_i $, which must then be empty.

However, in the aforementioned complette lattice, $\cap$ and $\wedge$ coincide on all sets but the emptyset. It is easy to see that in a complete lattice, the greatest lower bound of the emptyset is the lattice's maximum element. Therefore $\bigwedge_{i\in \emptyset}A_i = \bigwedge\emptyset = S$. Note that here, $\bigwedge$ does depend on $S$ so this makes complete sense.

However, in practice, since $\bigwedge$ and $\cap$ coincide so often, we ignore those subtleties and declare them to be equal, which leads to the abuse of notation $\bigcap_{i\in \emptyset}A_i = S$, which is what the other answers essentially tell you. But this is an abuse of notation (by the usual definitions etc.)

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    $\begingroup$ Huh? The intersection is given by a formula. Vacuous truth means that the intersection over the empty set is everything. What is wrong exactly? Yes, you can adopt the standard that the intersection is always taken from the union, thus ensuring the intersection over the empty set is empty again, but there is useful situations (most of them) where you want the empty intersection to be everything. $\endgroup$ – Asaf Karagila♦ Nov 10 '17 at 18:18
  • $\begingroup$ @AsafKaragila : surely $\{x\in S\mid \forall i\in I, x\in A_i\}$ cannot be a definition for $\cap_{i\in I}A_i$, because the $A_i$ are just sets and $\cap$ should (and actually can) be defined uniformly, because it is an operation that depends only on the sets in question, not on some other structure. Obviously, you can define $\cap_S$ in a sense, and denote it $\cap$, but why create a new definition when we have perfectly valid ones ? (And when $\cap$ being an abuse for $\wedge$ works well) $\endgroup$ – Max Nov 10 '17 at 21:24
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    $\begingroup$ Huh? I don't understand your issue. The intersection over a set $A$ is defined as $x\in\bigcap A\iff\forall a\in A:x\in a$. This is a uniform definition, just as uniform as $x\in\bigcup A\iff\exists a\in A:x\in a$. The only issue here is that you can prove that $\bigcap A$ is a set if and only if $A\neq\varnothing$; whereas you can prove that $\bigcup A$ is a set always, due to the axiom of union. $\endgroup$ – Asaf Karagila♦ Nov 10 '17 at 21:26
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    $\begingroup$ @AsafKaragila : yes precisely, but for the definition to always make sense, restricting $\cap A$ to $\cup A$ is an easy trick and doesn't cost much. Otherwise you end up with a "function" not defined everywhere. Moreover, with the definition you just gave, you must see that the answer to "What is $\cap \emptyset$ ?" cannot be "It's $S$ ! " $\endgroup$ – Max Nov 10 '17 at 21:28
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    $\begingroup$ Since the language of set theory only has $\in$, I don't see the issue with "partial operators". So the fact that $\bigcap\varnothing$ is a proper class is just peachy. As for the fact that it is "the universe", this is something mathematicians call context (maybe you've heard the term somewhere? It means the general settings in which we work at the moment). And it is easy to have a "universal class" which may or may not be a set, such that all sets in questions are subsets of the class (and maybe also elements of it, e.g. V), and then $\bigcap\varnothing$ is this universal class. $\endgroup$ – Asaf Karagila♦ Nov 10 '17 at 21:29
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$(x ∈ A_{\alpha} , \forall \alpha \in I )$ does not imply $( \exists \alpha \in I \text{ such that } x \in A_{\alpha} )$, since I may be $\emptyset$

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