Since complete metric spaces are constructed the same way that one would build up $\mathbb{R}$ from $\mathbb{Q}$, does that suggest that the least upper bound property is similarly inherited?
Asked
Active
Viewed 141 times
1 Answers
4
This question doesn't make any sense because to talk about a least upper bound you need an order and a general metric space doesn't have one.
J. De Ro
- 21,438
-
Is there some natural ordering for general metric spaces? – Stupid Questions Inc Jan 16 '20 at 12:09
-
Or let's say assuming the axiom of choice, does the least upper bound property hold for a complete metric space under the well-ordering guaranteed by the axiom of choice? – Stupid Questions Inc Jan 16 '20 at 12:12
-
You do realise that a well-ordening on the reals does not coincide with the usual order on $\mathbb{R}$, right? I see no reason why that would be true – J. De Ro Jan 16 '20 at 14:46