Proving the Least Upper Bound & Greatest Lower Bound Properties in Set A

[Tom1992]

prove that if A has the least upper bound property, then it also has the greatest lower bound property.

the least upper bound property means that if A has an upper bound, then it also has a least upper bound. the greatest lower bound property means that if A has a lower bound, then it also has a greatest lower bound..

the common proof is to simply show that the least upper bound of all the lower bounds of a subset B of A is equal to the greatest lower bound of B. i know about this proof, but i came up with my own proof (and my own lemma). can someone check if it is correct? i personally found no logical error. i typed it out in ms word (with capitilized sentences).

my idea was to first show that if < is an order relation, then > is also an order relation, then you can imagine what i did next (lack of glb would mean lack of lub, a contradiction).

 

[HallsofIvy]

prove that if A has the least upper bound property, then it also has the greatest lower bound property.
the least upper bound property means that if A has an upper bound, then it also has a least upper bound. the greatest lower bound property means that if A has a lower bound, then it also has a greatest lower bound..

the common proof is to simply show that the least upper bound of all the lower bounds of a subset B of A is equal to the greatest lower bound of B. i know about this proof, but i came up with my own proof (and my own lemma). can someone check if it is correct? i personally found no logical error. i typed it out in ms word (with capitilized sentences).

my idea was to first show that if < is an order relation, then > is also an order relation, then you can imagine what i did next (lack of glb would mean lack of lub, a contradiction).

I have absolutely no idea how you think anyone can check if what you did is correct when you expect us to "imagine" what you did!

Part of the problem may be reflected in your reference to A having the upper bound property. It makes no sense to talk about a set alone having the upper bound property: the set must have a specific order relaion on it.

Assuming that set A, with order relation <, has the least upper bound property, you cannot simply assume that A, with > order, also has the least upper bound property.

 

[Tom1992]

I have absolutely no idea how you think anyone can check if what you did is correct when you expect us to "imagine" what you did!

Part of the problem may be reflected in your reference to A having the upper bound property. It makes no sense to talk about a set alone having the upper bound property: the set must have a specific order relaion on it.

Assuming that set A, with order relation <, has the least upper bound property, you cannot simply assume that A, with > order, also has the least upper bound property.

did you read my proof in the ms word attachment? so my lemma proof is correct, but the main proof is wrong because A does not have the least upper bound property in the second order relation?

 

[Tom1992]

yes, you are correct! i failed to realize that the least upper bound property only applies to a SPECIFIC order relation--not all possible order relations. i didn't read the definition carefully. I'm sure the proof of my lemma is correct though. thanks a lot.