3
$\begingroup$

„A family of sets is pairwise disjoint or mutually disjoint if every two different sets in the family are disjoint.“ – from Wikipedia article "Disjoint sets"

What about the empty family of sets? Is it also pairwise disjoint?

I think, that the empty family of sets is pairwise disjoint, because statements of the form $\forall x \in \emptyset:\ldots$ are always true. Am I right?

$\endgroup$

2 Answers 2

4
$\begingroup$

Yes, you are right. It is vacuously true. Here's a more detailed explanation of why:

In math, either a statement is true, or its negation is true (but not both). That means, for example, either the statement (a) $\forall x \in \emptyset$, $x^{2} = 1$ or its negation, (b) $\exists x \in \emptyset$ such that $x^{2} \neq 1$, is true, and the other is false.

It's clear that statement (b) is false since $\exists x \in \emptyset$ is a false statement. So, since statement (b) is false, its negation, statement (a), must be true (it's called vacuously true).

$\endgroup$
1
  • $\begingroup$ "In math, either a statement is true, or its negation is true (but not both)" actually, this is only valid in classical logic, but it really doesn't matter for this question $\endgroup$
    – univalence
    Commented Apr 7, 2020 at 19:19
1
$\begingroup$

If $A$ is not a family of sets which are pairwise disjoint, then there exists $A_1,A_2\in A$ such that $A_1\neq A_2$ and $A_1\cap A_2\neq\varnothing$.

So... yes.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .