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Reading on Wikipedia, I found a definition of intersection based on the axiom of separation. Basically, to define $\bigcap A$, they demand that $A$ is not empty, take an element of $A$ that they call $E$ and say that $\bigcap A = \{c \in E: \forall D (D \in A \Rightarrow c \in D)\}$. This definition leaves $\bigcap \varnothing$ undefined.

What if instead of using $E$, we used the union of $A$? $\bigcap A = \{c \in \bigcup A: \forall D (D \in A \Rightarrow c \in D)\}$. In this case, when $A$ is empty, $\bigcup A$ is empty, $c \in \bigcup A$ is false and plugged in the axiom of separation it tells us that for any $c$, $c \notin \bigcap A$. So, $\bigcap \varnothing$ is simply $\varnothing$.

Can someone help me understand what is wrong in this reasoning? Thank you.

The curious amateur
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    This definition is used sometimes. However, it runs into issues. For example it is no longer true that if $A\subseteq B$, then $\cap B\subseteq A$, and it is no longer true that $\cap(A\cup B) = (\cap A)\cap (\cap B)$. – Arturo Magidin Feb 01 '24 at 22:38
  • @ArturoMagidin, if I wanted to prove $A \subseteq B \Rightarrow \bigcup B \subseteq A$, I would have to require that $B\neq \varnothing$. But so do you, because $\bigcap B$ is non-defined for you if $B=\varnothing$. Again, unless you get out of ZF and think in terms of proper class of all sets and stuff I confess I don't know much about. – The curious amateur Feb 01 '24 at 23:03
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    The property is (I had a typo) $A\subseteq B\implies \cap B\subseteq \cap A$. This holds for nonempty $A$. If you do not require nonempty $A$, then the problem arises with the proposed definition when $A$ is empty and $B$ is nonempty with nonempty intersection, as then you have $\cap B\neq\varnothing$ but $\cap A=\varnothing$. You have two choices in ZF: you can define the empty intersection to be empty, and add a bunch of caveats to standard theorems about unions and intersections; or you can exclude the empty intersection. It is usually (but not always) less of a hassle to do the latter. – Arturo Magidin Feb 01 '24 at 23:06
  • Note that the property @ArturoMagidin gave still holds if you use the same definition as they give on wikipedia for $\bigcap\varnothing$. Then you get that it is the collection of all sets (i.e. the universe). The only problem you might have with this is that the universe is a proper class. – Vincent Batens Feb 01 '24 at 23:16
  • @VincentBatens Right; that definition works in theories like GBN, but not in ZF. – Arturo Magidin Feb 01 '24 at 23:17
  • Can't you like just define class terms as terms (not as objects) in ZF just like analogously as how they defined class terms on page 4 of this (4.4). – Vincent Batens Feb 01 '24 at 23:23
  • I guess by defining that, you extend the language and you are probably technically not working in ZF anymore right? – Vincent Batens Feb 01 '24 at 23:28
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    Thanks Vincent. It'll take me a while to digest these class terms. It's fascinating to see what different people come up with. – The curious amateur Feb 01 '24 at 23:31
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    @VincentBatens: Indeed, that's not ZF any more. – Arturo Magidin Feb 01 '24 at 23:33
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    @VincentBatens I think of class terms as just being a different presentation of ZF. It is effectively the same as treating $\cap$ as a partial function where we just don't have any set equal to it if applied to $\emptyset.$ – spaceisdarkgreen Feb 01 '24 at 23:33
  • ye that's fair, in the end it indeed is the same – Vincent Batens Feb 01 '24 at 23:37

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The definition given on wikipedia is used as well for the empty set as you can see here (Nullary intersection).

Then $\bigcap\varnothing=V$ where $V$ denotes the universe because $(D\in\varnothing\Rightarrow c\in D)$ is trivially true for any set $c$ since $D\in\varnothing$ does not exist. (On wikipedia they limit themselves to this when working inside of some set $X$, but for example in the book Set Theory from Peter Koepke, they also say that $\bigcap \varnothing=V$ and everything works.)

The intuïtion (as it is also mentionned on wikipedia) is that if you take the intersection of less and less sets, the resulting set should get bigger and bigger, since $\bigcap U'\supseteq\bigcap U$ if $U'\subseteq U$.

Edit: I was actually (without realizing) working in NBG instead of ZF where you extend ZFC with class terms. In the end you choose definitions for edge cases such that the properties you like keep working for your edge cases.

Vincent Batens
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  • In ZF there can't be a universe, so I don't think that helps me that much. However, I am appreciating how, intuitively, the intersection of something small should be something big, so saying that the intersection of the empty set is empty is a bit like saying that 7 divided by 0 is 0. – The curious amateur Feb 01 '24 at 23:08