This would be true if there were such thing as an “unordered pair”
-
-
-
There is.
-
An unordered pair is just an equivalence class of ordered pairs.
-
Try writing the letters A and B without writing them in a particular order
-
The symbol is not the referent.
-
What is the relation between the symbol and the referent?
End of conversation
New conversation -
-
-
Disjoint union corresponds to addition and works for any two sets. Swapping them yields an equal set. Set product requires one set to be designated as an "indexing set". Swapping them is set-isomorphic but not equal.
-
that makes no sense... first of all, disjoint union is also only commutative up to isomorphism; second of all, you're only saying that disjoint union is commutative, which is exactly what OP says about *multiplication*
-
Sorry, I meant "union of disjoint sets" not "disjoint union". Multiplication of sets is not commutative. Multiplication of set equivalence classes is.
-
ok, but nonetheless, you're only saying that addition is commutative and multiplication is commutative up to a weaker equivalence. that doesn't really correspond to the tweet.
-
Commutative ⊨ Unordered ~Commutative ⊨ Ordered
-
like i said, though, that doesnt make sense when the original tweet literally said "commutative" for that matter, it also said "numbers", not "sets", and if you wanna argue that it meant ordinal or cardinal numbers, well, those are already considered up to a weaker equivalence
-
The original tweet is incorrect under a strict reading. The principle of explosion allows no insights to be gained from an incorrect statement. Happy?
- 1 more reply
New conversation -
-
Why is this not an artifact of a definition? Afraid I don’t get the point
Thanks. Twitter will use this to make your timeline better. UndoUndo
-
Loading seems to be taking a while.
Twitter may be over capacity or experiencing a momentary hiccup. Try again or visit Twitter Status for more information.