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I had to prove the existence of the intersection of a nonempty set S, and my first approach was the following:

Let $S$ be an arbitrary set. By the Axiom of Union, the set $ \cup S$ exists; we can then apply the Axiom of Extensionality and to talk about the set $I=\{x \in \cup S: (\forall T \in S) x \in T \}$.

This set $I$ seemed to be the good candidate for being the intersection, but then I realized I never used the hypothesis of $S$ being non-vacuous ! Moreover, with this "proof" I'll have that $\cap \emptyset$ will be a subset of $\emptyset$, then equal to the empty set, but I know very well that's false.

I already figured out a right proof of my initial problem:

let $S$ be a non-empty set; let then A be an element of S. We can then apply the Axiom of Extensionality and to talk about the set $I=\{x \in A: (\forall T \in S) x \in T \}$, and this set is what we wanted to be something called "the intersection of the non-vacuous set $S$ ".

but now I'm curious about knowing what's wrong with my first tricky argument (just please don't be mean with me, I'm aware it can be a silly mistake).

William Elliot
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Amelian
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  • You may find this useful: https://math.stackexchange.com/questions/370188/empty-intersection-and-empty-union https://math.stackexchange.com/questions/6613/unary-intersection-of-the-empty-set – Jean-Claude Arbaut Jan 17 '21 at 17:41
  • If $S$ is empty, $\bigcup S= \emptyset$ – Mauro ALLEGRANZA Jan 17 '21 at 17:41
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    There is an issue with $\bigcap \emptyset$: if we apply the def of $\bigcap$ to $\emptyset$ we get $V$: $\bigcap S = { x \mid (\forall A) (A \in S \to x \in A) }$. If $S$ has no elements, thus the specifying condition is always true and every $x$ will be a member of $\bigcap S$. – Mauro ALLEGRANZA Jan 17 '21 at 17:54
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    Your "proof" assumes that the intersection of the family $S$ must be a subset of the union, $\bigcup S$. That follows easily from the dfinitions for all nonempty $S$, but not for empty $S$. – Andreas Blass Jan 17 '21 at 18:11

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We understand intuitively that a set B is the intersection of a family $S$ iff for all $x$ we have that $$ x \in B \iff (\forall X \in S)x \in X. $$ It's obvious that we can draw $(\forall X \in S)x \in X$ from $x \in \{x \in \bigcup S : (\forall X \in S)x \in X\}$.

If $(\forall X \in S)x \in X$ and $S \neq \varnothing$, then there is a $B \in S$. So, $(\exists X \in S)x \in X$ and thus $x \in \bigcup S$. We conclude that $x \in \{x \in \bigcup S : (\forall X \in S)x \in X\}$.

What if $S = \varnothing$? Then the last reasoning now will be broken because we can't choose a concrete element of $S$.

Technically your first definition is valid, but the second one is simpler and more transparent with the use of $S \neq \varnothing$.

Alejandro
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