The Completeness Axiom only asserts something about sets that are bounded above.
2007-02-22 16:38:12 · 4 answers · asked by Kelvin K 2 in Science & Mathematics ➔ Mathematics
WOW! Dude, you must be really smart because to me it sounds like another language!
2007-02-22 16:50:38 · answer #1 · answered by ♥Tawnya♥ 4 · 0⤊ 2⤋
I read a proof on another web page, and it used the Completeness Axiom. Basically it assumed an ordered set of numbers in Reals. Then you can take the negative of those and obviously exactly reverse the order. By the Completeness Axiom, there exists a supremum of that reversed set. So then obviously, if you take the negative of those numbers, then there must also exist an infimum.
2007-02-22 17:33:08 · answer #2 · answered by David S 4 · 1⤊ 0⤋
I study a suggestion on yet another internet website, and it used the Completeness Axiom. surely it assumed an ordered set of numbers in Reals. then you definately can take the unfavorable of those and obviously precisely opposite the order. by the Completeness Axiom, there exists a supremum of that reversed set. So then obviously, when you're taking the unfavorable of those numbers, then there also ought to exist an infimum.
2016-12-04 20:06:25 · answer #3 · answered by ? 4 · 0⤊ 0⤋
Any ordered set with the least upper bound property also has the greatest lower bound property.
2007-02-22 16:57:38 · answer #4 · answered by a_liberal_economist 3 · 0⤊ 0⤋