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?

$\endgroup$
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).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ "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 $\endgroup$ – Mega Man Apr 7 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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.