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"...which makes the proof wrong, since not all metric spaces have obvious orderings on their elements..." Your reasoning here is wrong. The "a" and "b" come f
by mturmon 14y ago
"...which makes the proof wrong, since not all metric spaces have obvious orderings on their elements..."
Your reasoning here is wrong. The "a" and "b" come from the range space of the metric, which by definition associates a pair of points in the metric space (unnamed, but call it X) with a nonnegative real number.
In short, "a" and "b" are in R, not in the original metric space X, so it's legal to take the lesser of a and b.
- AimHere 14y agoOops yes. I see that now. I told you I was rusty!