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In what sensible way can we define the union and intersection of an empty family of sets? How come empty intersection of empty sets become the whole space ?

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  • $\begingroup$ Great question, +1 $\endgroup$ – Rustyn Apr 5 '14 at 9:17
  • $\begingroup$ @Rustyn: Great question? Arguable. Question that appears periodically on this website, and should be closed as a duplicate? Definitely. $\endgroup$ – Asaf Karagila Apr 5 '14 at 14:38
  • $\begingroup$ See also math.stackexchange.com/questions/716148/…, and many links which appear in both threads. $\endgroup$ – Asaf Karagila Apr 5 '14 at 14:42
  • $\begingroup$ Huh, wasn't aware that it was a duplicate. @asaf $\endgroup$ – Rustyn Apr 5 '14 at 19:54
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The usual definition of intersection is $x \in \bigcap_{i \in I} A_i$ if and only if $(\forall i \in I)(x \in A_i)$.

If $I$ is empty, every $x$ vacuously satisfies $(\forall i \in I)(x \in A_i)$.


In some regard, this makes sense. In general, if $I \subseteq J$ and $\{A_i : i \in J\}$ is a collection of sets, you should expect $\bigcap_{i \in J} A_i \subseteq \bigcup_{i \in I} A_i$ from any reasonable definition.

You can arbitrarily define the empty intersection to be some set (in the sense of set theory), but it will not have the property of the previous paragraph with respect to every collection $\{A_i : i \in I\}$. Although the collection of all set is not a set (in the sense of set theory), it however informally does satisfy this property.

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Suppose that $x$ does not belong to the empty intersenction. Then, there is a set $A$ such that $x\not\in A$, but this is a contradiction, since there is no set $A$ at all.

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The union of an empty family of sets is empty itself, and well-defined. An intersection of an empty intersection only makes sense if you are working under the convention that you are in some universe or set $U$. If the $A_i$ are subsets of $U$ then you can define $$\cap_{i\in\emptyset}A_{i}:=U$$ If $u\in U$ then $u\in A_{i}$ is vacuously true for each $i\in\emptyset$

(no $i\in\emptyset$ exists with $u\notin A_i$. Even stronger: no $i$ exists with $i\in\emptyset$).

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