3 ms·
Yes, this looks like an error. You don't want unordered pairs, which do not form a vector space, but an isomorphism of under order permutation of the terms. Rou
by knappa 4y ago
Yes, this looks like an error. You don't want unordered pairs, which do not form a vector space, but an isomorphism of under order permutation of the terms. Roughly speaking, V⊕W is not the same as W⊕V, but if you swap the order everywhere then math won't notice.
- deleted 4y ago[deleted]
- bmacho 4y agoI think it's correct, and also seems pretty deliberately. I also think that his way is inferior to the common way (I can't think of 1 single advantage of it, really).