Discussion in 'Physics & Math' started by arfa brane, Feb 25, 2017.

1. ### arfa branecall me arfValued Senior Member

Messages:
5,260
If {a,b} ∩ {b,c} = {b}, and {a,a} ∩ {b,c} = ∅, what is {a,a} ∩ {a,c}?

1) {a,a}
2) {a}

3. ### QuarkHeadRemedial Math StudentValued Senior Member

Messages:
1,492
$\{a,a\}$ is not a set (no element can appear more than once in a well-formed set)

rpenner likes this.

5. ### arfa branecall me arfValued Senior Member

Messages:
5,260
Ok. Nonetheless I recall being told by a professor that the set {a} is distinct from the set {a,a}. But apparently {a,a} isn't a set.
If I accept that then I need to accept that {a,a} cannot be the union or intersection of other sets, nor can it be the complement of a set . . .?

7. ### rpennerFully WiredStaff Member

Messages:
4,833
According to the axiom of pairing, if X and Y are set-valued variables, then there is a set W such that Z ∈ W means Z = X OR Z = Y. In other words, W = { X, Y }.
According to the axiom of extensionality, if X and Y are set valued variables then if for every Z, Z ∈ X was true exactly whenever Z ∈ Y, then X is the same set as Y.
Therefore {a, a} = {a} and {a,a} ∩ {a,c} = {a}. To write {a, a} as a technically correct solution while ignoring {a} would be a flaw in your understanding of sets.

https://en.wikipedia.org/wiki/Axiom_of_pairing
https://en.wikipedia.org/wiki/Axiom_of_extensionality

{a} and {a, a} are distinct in the mathematical realm of multisets, but that is not the topic you asked about.

In mulitsets, {a,a} ∩ {a,c} = {a} because 1 = min(2,1) and 0 = min(0,1).

https://en.wikipedia.org/wiki/Multiset

8. ### arfa branecall me arfValued Senior Member

Messages:
5,260
Ok, thanks for the expl.

All I remember being told is that {a,a} is not the same "kind of object" as {a}.

Another thing I still haven't quite grasped is about Cartesian products of sets.
Any set A × ∅ = ∅. But isn't the product a set of duples: {(a,∅), . . . }, how is that mathematically = ∅?

9. ### rpennerFully WiredStaff Member

Messages:
4,833
Your example is misleading, because there is literally no element which can form the second part of that ordered pair.

10. ### QuarkHeadRemedial Math StudentValued Senior Member

Messages:
1,492
Because in set theory, the emptyset symbol is often (always?) used to denote non-existence. For example writing $A \cap B=\emptyset$ means "there does not exist any element in the intersection of A and B"

So $A \times \emptyset= \emptyset$ is perhaps technically correct, since, as rpenner says, there exists no ordered pair which can serve as an element in this Cartesian product.

But it is not helpful. Far from it.

11. ### arfa branecall me arfValued Senior Member

Messages:
5,260
I wonder what is helpful then, in trying to understand the equality A × ∅ = ∅.

Since it is something you encounter, and the explanation or proof if you will, is that there are ordered pairs (a,b) such that a ∈ A and b ∈ ∅. But there is no b ∈ ∅. So there are no ordered pairs and the product is empty.

Is the 'axiom of pairing' useful (or helpful) here?

12. ### rpennerFully WiredStaff Member

Messages:
4,833
The axiom of pairing says that if A is a set and B is a set then U = {A, B} is a set. So V= { A }, W= { B }, X = { ∅ }, Y= { A, ∅ }, and Z = { B, ∅ } are also also sets by the axiom of pairing. But that means ∅ is an element of { ∅ }, so { ∅ } is not the empty set just like V≠A and W≠B.

So I don't think it's useful, since the properties of ∅ come from it being empty.

card ∅ = 0
card X = 1
card U = 2
card ∅×∅ = 0
card ∅×X = 0
card ∅×U = 0
card X×X = 1
card X×U = 2
card U×U = 4
card A×B = (card A)×(card B) (when A and B are finite sets)