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Let's say A=empty set and every element of A also a set, what is the intersection of all elements of A?

real answer:?

possible answer: universal set (I found it via using Russel paradox. And then I thought backward.)

moccho
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2 Answers2

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Every element of the empty set is a set, no need to assume that. But the intersection of the elements of the empty set is undefined; the definition of intersections applies only to non-empty families of sets.

(Why is it undefined? If we try to apply the definition, with $I$ being the intersection of the elements of the empty set, we get $$x\in I\iff\forall y\in \emptyset\,\,\,\, x\in y.$$But that says that every $x$ is in $I$, and there is no "universal set" in modern set theory)

David C. Ullrich
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Proof of $\forall i, X_i\subseteq X$, then $\bigcap_{i\in\emptyset}X_i=X$ is like this. However, to avoid Russell's paradox, it must first be implicitly assumed that all actions take place under a particular set(in this case, $X$).

For all $x\in X$, $i\in\emptyset \Rightarrow\bigcap_{i\in\emptyset} x\in X_i$ holds, because it is vacuous truth, since antecedent $i\in\emptyset$ is false.

MH.Lee
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    Your proof shows $X\subseteq \bigcap_{i\in\varnothing }X_i$. For equality, you need to show the other inclusion. But the other inclusion is not true unless you define $\bigcap$ in such a way that only elements of $X$ are considered. To illustrate the issue, note that all sets in the empty family are subsets of $\mathbb{N}$, so your assertion is that the empty intersection is equal to $\mathbb{N}$. But also all sets in the empty family are subsets of $\mathbb{R}$. Is the empty intersection also equal to $\mathbb{R}$? – Alex Kruckman Aug 21 '22 at 13:29
  • @AlexKruckman I added the corresponding explanation. – MH.Lee Aug 21 '22 at 13:47
  • that explanation doesn't help - what you say is simply false. How does saying "all actions take place under a particular set invalidate the counterexamples you've been given? – David C. Ullrich Aug 21 '22 at 13:49