6 ms·
Why wouldn't the following ordering work? {(x,y) | (|x| ≤ |y|) & ((|x| = |y| & x ≠ y) ⇒ x < 0 < y)}
by mrmyers 9y ago
Why wouldn't the following ordering work?
{(x,y) | (|x| ≤ |y|) & ((|x| = |y| & x ≠ y) ⇒ x < 0 < y)}
- v64 9y agoWith your ordering, what is the least element of the open interval (0,1)?
- mrmyers 9y agoAh, right. Makes sense.
- nerdponx 9y agoWhat if you order the absolute values of the differences from the midpoint of the interval, where the midpoint is supremum - infimum if bounded, or zero if unbounded? Obviously this fails for closed/open sets like [a,infinity), but wouldn't it work for fully open sets?
- v64 9y agoUltimately, the point becomes moot because we focus too much on "normal" sets. Such a definition does nothing to help us answer questions like, what is the least element of the irrationals?