If a collection on $\Omega$ is closed under finite intersections, then can we conclude that it contains $\Omega$?

Question

Here an example that illustrates my question.

A $\pi$-system $\mathcal P$ on set $\Omega$ is defined here to be a collection that is not empty and satisfies the rule $A,B\in\mathcal P\implies A\cap B\in\mathcal P$.

Then they go on by saying that the collection is closed under finite intersections.

But wait a minute... Isn't $\varnothing$ a finite set?

Interpreting the empty intersection as $\Omega$, this can make people think that $\Omega\in\mathcal P$, which is not a consequence of the definition.

($\varnothing\subseteq\mathcal P$ and it is vacuously true that $\omega\in P$ for every $\omega\in\Omega$ and every $P\in\varnothing$)

Personally I would rather go for "closed under binary intersection" keeping in mind of course that this implies "closed under non-empty finite intersections".

Thanks in advance for sharing your thoughts on this.

Answer 1

They don't mean intersections of finite sets, they mean intersections of finite numbers of sets. For instance: if A,B, and C are sets in P, then $A\cap B\cap C\in P$ In general, this is true because, the base case follows from the definition of P and if $\cap$ {$A_1,A_2,....A_k$} $\in P$, for $k\in \mathbb N$, and $A_{k+1}$ is any set, ($\cap$ {$A_1,A_2,....A_k$})$\cap A_{k+1}$ is just an intersection of 2 sets, which we know is in P by the definition of P. Thus, by induction on the number of sets in an intersection P is closed under the intersection of any finite number n of sets: i.e. closed under finite intersection.

Also, from the information you provided, it doesn't even follow that the empty set has to belong to a given pi system of a set omega, although there are surely some pi systems that do; for instance, the pi system {O}. Clearly this is a pi system of any set omega since it's a non-empty collection of subsets of omega that is closed under intersections of any finite number of sets within it.

Answer 2

Okay. I see what you were asking now. This is a really cool question actually, I worked out something really interesting at my desk; and I'm also including my thought process that led me to my answer at the bottom of the page.

It turns out that it is not clear how to define $\cap${ } so that it is a set which actually exists. I have never seen, anywhere, a definition of the intersection of a set of zero sets; and now I've discovered the reason:

Lets consider the most natural and tempting definition of $\cap$ { }:

$$\cap \textrm {{ }}= \textrm{{y:$\forall x\in \textrm{{ }}, y\in x$}}$$

In ZFC, this is a class. What remains to be found is whether or not it's proper.

By the law of excluded middle in classical logic,

$$\textrm{(1) }\exists\cap \textrm{{ }}\oplus \nexists\cap \textrm{{ }}$$

Also, using that same law again,

$$\textrm{(2) } \exists\cap \textrm{{ }}\rightarrow \cap \textrm{{ }}=\varnothing\oplus\cap \textrm{{ }}\ne\varnothing$$

Where $\oplus$ is the "exlusive or" symbol.

Supposing the former in (1), also suppose the former in the predicate of (2). Then,

$$\forall y, y\notin \cap \textrm{{ }}$$ By our definition of $\cap \textrm{{ }}$, it is necessary that

$$\exists x\in\textrm{{ }}\ni y\notin x$$

However, this is a contradiction because{ } is empty. This would force us to evaluate other cases and such. A few minute's of thought, though, about our definition of $\cap \textrm{{ }}$ reveals just how large of a set this set would have to be:

I claim that this intersection contains every set. $$\textrm{Let y be a set. Now suppose $x\in y$. This is obviously false, so any predicate to it evaluates as true. Thus $y\in\cap \textrm{{ }}$.} $$

But from the notion of a class in ZF (or ZFC), $\cap \textrm{{ }}$ is an aggregation of sets. Thus, we have,

$$y\in \cap \textrm{{ }}\leftrightarrow y\in\textrm{"The class of all sets"}.$$

The latter class is known to be a proper class, that is, a thing that is a class but not a set.

There's your answer: Unless we come up with a better definition of $\cap \textrm{{ }}$, we need to make a stipulation that "it be closed for any finite intersection of 1 or more;" because the standard definition of intersection does not provide a set which exists." This all stems from the fact that the set of all sets is not actually a set. This is the result of one of the classical paradoxes that led to the axiomatization of set theory.

If this idea troubles you, let the set of all sets be S. Consider the subset A={x$\in$S:x$\notin$x} (This is the set in Russell's paradox, which Bertrand Russell provided as a counterexample to Frege's axiom of abstraction). Clearly,

$$ A\in A \oplus A\notin A$$

Suppose the first. Then $A\notin A$. This is a contradiction. Suppose the latter, then $A\in A$. This is a contradiction too. So, A cannot exist. The only thing that went wrong with our proof was our supposition that the set of all sets is a set.

-Adam V. Nease

Answer 3

I agree that for maximal naturality and elegance we should demand that $\Omega$ is in every $\pi$-system.

However, the notion of a $\pi$-system is already not a particularly natural one and is only really used in the $\pi-\lambda$ theorem that says that any $\lambda$-system contains a $\pi$-system then it also contains the $\sigma$-algebra generated by that $\pi$-system. Since $\sigma$-algebras and $\lambda$-systems already have to contain $\Omega$ this theorem would only be made weaker by demanding that $\Omega$ be in the $\pi$-system.

As it is we have made the $\pi-\lambda$ theorem slightly stronger, and at not much expense of elegance, since the notion of a $\pi$-system would is pretty ugly anyway.