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$\def\C{\mathbb{C}} \def\R{\mathbb{R}}$

First, let's review the concept of an empty family (a family of sets with an empty index set):

Definition. Let $X$ be a set. A mapping $A:\emptyset \to 2^X$ is called an empty family, denoted by $\left\{A_i\right\}_{i \in \emptyset}$.

The following fact is well known:

Proposition. $\bigcap_{i\in\emptyset} A_i=X$.

The standard explanation for this is as follows:

Proof. For $x\in X$, the statement $x\in\bigcap_{i\in\emptyset} A_i$ is equivalent to $\forall i(i\in\emptyset\implies x\in A_i)$, which is always true due to the so-called "vacuous truth." $\square$

Now, let's set $X=\R$. According to the above proposition, we have $\bigcap_{i\in\emptyset} A_i=\R$.

Next, let's set $X=\C$. According to the above proposition, we have $\bigcap_{i\in\emptyset} A_i=\C$.

Therefore, we obtain $\R=\bigcap_{i\in\emptyset} A_i=\C$, implying $\R=\C$. Where is the mistake?

My personal suspicion: In this case, the object "$A_i$" cannot be considered in the first place, so it is questionable whether the expression "$x\in A_i$" is a well-defined logical formula.

Edit: Thank you for your answers. My understanding is wrote in the reply to Dan Doel's answer, if it is wrong please correct it. And how do you think of "My personal suspicion"? Some authors claim $\left\{A_i\right\}_{i \in \emptyset}$ is a family of subsets of $X$ (and the proposition hold), but I feel weird by this suspicion.

user682141
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    It is not so much that $\bigcap_{i\in\emptyset}A_i = \Bbb R$... it is that in the context of $X=\Bbb R$ we have $\bigcap_{i\in\emptyset}A_i = \Bbb R$. The context must be kept. – JMoravitz Jun 05 '24 at 15:34
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    Your proof is incorrect. It's not only for $x \in X$ that $\forall i(i\in\emptyset\implies x\in A_i)$, it's for any $x$ at all. So by this logic $\bigcap_{i \in \emptyset} A_i$ is not just $X$, it is "everything" (which, of course, is not a set at all). In order to have a sensible definition of $\bigcap_{i \in \theta} A_i$, you must specify that you're restricting it to members of $X$. And then, of course, the result depends on which $X$ you're using. – Robert Israel Jun 05 '24 at 15:36
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    This is why types matter. The intersection of $A : I \to \mathcal{P}(X)$ is a subset of $X$. Your two $A$s have different types and therefore their intersections cannot meaningfully be compared for equality. – Naïm Camille Favier Jun 05 '24 at 15:47
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    Isn't it the case that $\bigcap_{i\in\varnothing} A_i$ doesn't exists over $ZFC$? – Antares Jun 05 '24 at 16:06
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    Right, it's a syntactic shortcut that implicitly depends on $X$. – Naïm Camille Favier Jun 05 '24 at 16:09
  • If you "set" X=R=C, everything is true. – philipxy Jun 06 '24 at 01:56
  • @Robert Israel Thank you for your comment. I would like to hear your opinion on my reply to Dan Doel. In conclusion, my understanding is that If my personal suspicion is unfounded, the definition of $\bigcap_{i\in\emptyset} A_i :={x \mid \forall i\in\emptyset, x \in A_i}$ holds. If my suspicion is valid, then we define $\bigcap_{i\in\emptyset} A_i := U_{\rm{set}}$, where $U_{\rm{set}}$ is the class of all sets. In this case, for any set $X$, we have $(\bigcap_{i\in\emptyset} A_i)X = (\bigcap{i\in\emptyset} A_i) \cap X$ (for the def of the LH, see my reply to Dan Doel), and this is a set. – user682141 Jun 08 '24 at 03:58

4 Answers4

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Here's where I suspect your confusion lies.

In ZF(C), every empty family $\emptyset → 2^X$ is exactly the same set, the empty set. Therefore it seems like its intersection should be some particular set, and can only be one particular set.

However, $\bigcap_{i\in I}A_i$ is not a real operation in ZFC. Its meaning must be defined. Similarly $x \in \bigcap_{i\in I}A_i$ isn't a real ZF(C) proposition, so, you must explain what it means in terms of the latter.

This transformation is syntactic, and it is not uniform in just the family, making the result a function of what the family actually is as a set (because it cannot be as Robert Israel mentions in the comments). It involves the type ascribed to the family, which includes $X$ (as Naïm Favier mentions in the comments).

So, the translation of:

$$x \in \bigcap_{i\in \emptyset}A_i$$

is defined to be something like:

$$x \in \{x \in X : ∀i\in\emptyset. x \in A_i\}$$

The $x \in X$ part is necessary because there is no universal set for the empty intersection to be; it must be limited by considering subsets of some 'largest' set. And the choice of largest set is a function of the type ascribed to the empty family, not its semantic representation.

When you do this, it no longer appears that $ℝ = ℂ$. The sets are not syntactically identical, because different intersections are involved (one over $ℝ$ and one over $ℂ$).

Dan Doel
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    I think it is misleading to talk about types in the context of ZF and ZFC, because they are not typed systems. What you are saying is that $\bigcap A_i$ is an informal notation that has an implicit parameter - some set $X$ that contains each $A_i$ - that has to be inferred from the context. – Rob Arthan Jun 05 '24 at 21:26
  • I see, the "proof" of the proposition is actually a proof in a set theory (one larger than ZFC) that $\bigcap_{i\in\emptyset} A_i$ is the class of all sets, right? However, the latter part doesn't quite make sense to me. As you are saying, now $A=\emptyset$, so $\bigcap_{i\in\emptyset} A_i$ should be an object that does not depend on $X$. But your definition of $\bigcap_{i\in\emptyset} A_i$ depends on $X$. For example if the notation includes $X$ as in $(\bigcap_{i\in\emptyset} A_i)X$, it makes sense, but saying that is a definition of $\bigcap{i\in\emptyset} A_i$ is strange. – user682141 Jun 07 '24 at 12:51
  • Furthermore, it is questionable whether the logical formula $x\in A_i$ that appears in your definition of $\bigcap_{i\in\emptyset} A_i$ is well-defined by the same reason of "my personal suspicion". – user682141 Jun 07 '24 at 12:51
  • I don't think it's at all uncommon to leave a parameter implicit in a situation like this. You have already stated that the family is supposed to be interpreted as $\emptyset → 2^X$. That is enough context to guess that the maximal set should be $X$, and leave it out of the notation. You can include it some way if you like, but other people may not. I wouldn't even expect an actual formal language to require it, because if you do treat that information like a type (despite 'types' not being intrinsic concepts for ZFC), a computer can likely infer the right answer. – Dan Doel Jun 07 '24 at 15:54
  • By 'actual formal language,' I mean something you could actually write and have verified on a computer, but which has all these informal aspects like actual set forming operators and whatnot. ZFC doesn't have any of that. The only actual new symbol in ZFC is $\in$, so all of the usual, informal set formation operations, and notation like functions, families, etc. have to be decoded somehow. – Dan Doel Jun 07 '24 at 16:02
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$\cap_{i\in\emptyset} A_i = (\cup_{i\in\emptyset}A_i^C)^C = \emptyset^C = X$

So this can be thought of as a consequence of the fact hat "The complement of $S$" is not a function merely of $S$, but of what set $S$ is considered to be living in.

In $\mathbb R$, $\emptyset^C= \mathbb R$.

In $\mathbb C$, $\emptyset^C= \mathbb C$.

So if we argue $\mathbb R = \emptyset^C= \mathbb C$, we get $\mathbb R = \mathbb C$. But clearly $\emptyset^C$ is referring to different things in these two equations.

Acccumulation
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There is not much of a mystery is here. There is a similar situation with regard the infimum over an empty set of real numbers. This infimum is $+\infty$, whereas the supremum is $-\infty$. Similarly here, the empty intersection should be the whole space, so if you are talking about sets of real numbers you would get $\mathbb R$. In ZF you can't talk about an empty intersection in the abstract without specifying what the "whole space" is, because that would be quantifying over the set of all sets, which does not exist.

Mikhail Katz
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In the end $\bigcap_{A\in \emptyset}$ (and other definitions like that) is just a special definition that has, in reality, nothing to do with $\bigcap_{A\in M}$ for $M$ non-empty. Instead, it is defined solely to save somebody a few lines in a proof, where it would otherwise have to be handled as a special case.

So this is one of the rare cases in math where understanding the definition has as much value as just memorizing it.

ConnFus
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