Not understanding a claim of Bourbaki about inverse limits in "Theory of Sets"

Question

In Chapter 7 ("Inverse limits and direct limits"), subchapter 4 ("Conditions for an inverse limit to be non-empty"), Bourbaki lets $(E_\alpha)_{\alpha \in I}$ be a projective system of sets with connecting maps $(f_{\alpha \beta}) _{\alpha, \beta \in I}$, and for each $\alpha \in I$ he lets $\mathfrak S_\alpha$ be a set of subsets of $E_\alpha$ with the property that "every intersection of sets belonging to $\mathfrak S_\alpha$ also belongs to $\mathfrak S_\alpha$". And then he weirdly claims that

it follows in particular (by considering the intersection of the empty family) that $E_\alpha \in \mathfrak S_\alpha$.

Why should this happen? (This fact is used later in the proof of step 4 of Theorem 1, so it is not a mistype.) It is easy to construct a $\mathfrak S_\alpha$ by choosing $U, V \subsetneq E_\alpha$ and letting $\mathfrak S_\alpha = \{U, V, U \cap V\}$, so that $E_\alpha \notin \mathfrak S_\alpha$ in this example, contradicting Bourbaki.

(By adding one more condition, it seems that Bourbaki wants the $\mathfrak S_\alpha$ to mimick the set of compact subsets of $E_\alpha$ that would exist in a topological setting.)

Answer 1

This is one of those statements which are true vacuously:

• The empty intersection of subsets of a set $A$ is the whole $A$.
• The empty union of subsets of $A$ is an empty set.

In other words: if $\emptyset=\mathcal F\subseteq\mathcal P(A)$ is the empty family of subsets of $A$, then:

$$\bigcap_{X\in\mathcal F}X=A$$

and

$$\bigcup_{X\in\mathcal F}X=\emptyset$$

This is, for example, because $x\in\bigcap_{X\in\mathcal F}X\Longleftrightarrow(\forall X\in\emptyset)x\in X$, which is true for all $x\in A$, again vacuously. ("If you can find $X\in\emptyset$ such that $x\not\in X$, that would be a counterexample - but you can't!") You can similarly justify the second identity.

Note also, if the second identity ("empty union is empty") is easier to "swallow", then you can easily derive the first identity from it using DeMorgan's laws:

$$\bigcap_{X\in\emptyset}X=\left(\bigcup_{X\in\emptyset}X^c\right)^c=\emptyset^c=A$$

Answer 2

Remember that the intersection of the empty family consists of everything: $$\bigcap\emptyset=\{x: \forall a\in\emptyset(x\in a)\}=\{x: x=x\}.$$ Now of course there isn't actually a universal set, so this is an abuse of terminology. When we speak of "the intersection of the empty family," we really mean this in the context of some universal set: writing "$\bigcap \mathcal{A}_U$" for "$\{x\in U: \forall a\in\mathcal{A}(x\in a)\}$," we have $$\bigcap\emptyset_U=\{x\in U: \forall a\in\emptyset(x\in a)\}=U.$$

(EDIT: Why is $\bigcap\emptyset_U=U$? Well, take $x\in U$; I claim $x\in\bigcap \emptyset_U$. Otherwise, we would have to have some $X\in\emptyset$ with $x\not\in X$, but there is no $X\in\emptyset$ at all. "$x\in U\implies x\in\bigcap\emptyset_U$" is vacuously true in exactly the same way, but with a little more abstraction, as the sentence "Every flying purple dinosaur is a neo-Kantian" is true. Universal quantification over the emptyset yields truth always, and existential quantification over the emptyset yields falsity always.)

The choice of universal set when performing this operation is almost always omitted (and the notation above, or anything like it, never used in my experience) - in my opinion, sadly - on the grounds that it's always clear from context; in this case, it's $E_\alpha.$ Note that "universal set" here is a misnomer, since we're considering lots of different $E_\alpha$s; a better word would be "context," but I feel that the term "universal set" is useful when thinking about what "the empty intersection" should be, at least at first.

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EDIT: This is how I would write Bourbaki's definition with a bit more precision:

Suppose $\mathfrak{S}_\alpha$ is a collection of subsets of $E_\alpha$ such that for all $X\subseteq\mathfrak{S}_\alpha$, we have $$\bigcap X_{E_\alpha}\in\mathfrak{S}_\alpha.$$

If we take $X=\emptyset$, then we have $\bigcap X_{E_\alpha}=E_\alpha$ by the reasoning above; since $\emptyset\subseteq\mathfrak{S}_\alpha$ regarldess of what $\mathfrak{S}_\alpha$ is, we get $E_\alpha\in\mathfrak{S}_\alpha$ as desired.