Let A be a set with elements {a, b, c, ...}. With these elements we can make sets of pairs, e.g., {a, b}, {a, c}, {b, c}, ... These pairs are not ordered, i.e., {a, b} = {b, a}, because by definition sets which have all the same elements are equal.

How can we make ordered pairs, say <a, b>, such that <a, b> ≠ <b, a>?

Moderator Note

This (along with your previous thread on axioms) doesn’t appear to be in the brain teaser/puzzle category. Moved.

10 minutes ago, swansont said:

Moderator Note

This (along with your previous thread on axioms) doesn’t appear to be in the brain teaser/puzzle category. Moved.

You are right. My issue with putting it in any other category is that then it appears as if I'm asking for help with this question, which I'm not.

23 minutes ago, Genady said:

You are right. My issue with putting it in any other category is that then it appears as if I'm asking for help with this question, which I'm not.

Would a “quiz” (or other) disclaimer/tag in the title be appropriate/useful?

1 minute ago, swansont said:

Would a “quiz” (or other) disclaimer/tag in the title be appropriate/useful?

If you think it makes it clear, fine with me.

Would you add something like that to the title of this thread?

4 hours ago, Genady said:

Let A be a set with elements {a, b, c, ...}. With these elements we can make sets of pairs, e.g., {a, b}, {a, c}, {b, c}, ... These pairs are not ordered, i.e., {a, b} = {b, a}, because by definition sets which have all the same elements are equal.

How can we make ordered pairs, say <a, b>, such that <a, b> ≠ <b, a>?

{{a}, b} ≠ {{b}, a}

8 minutes ago, KJW said:

{{a}, b} ≠ {{b}, a}

I thought of {a, {a, b}}, but your solution should work, too.

Both solutions are different subsets of [math]\mathcal {P}(A\cup \mathcal {P}(A))[/math].

If you want ordered pairs, you need the Cartesin product a A with itself.

The pairing subsets in the product space then provide one way to define a relation.

Edited 19 hours ago19 hr by studiot

37 minutes ago, studiot said:

If you want ordered pairs, you need the Cartesin product a A with itself.

The pairing subsets in the product space then provide one way to define a relation.

That's correct. The question is then, how to define the Cartesian product AxB so that AxB≠BxA.