I'm dumb and can't figure out how to proceed on this proof

>>8987251
The lower bounds of B have no relation to the upper bounds of L in the assumptions. It's easy to find a subset of B not in L that contains the greatest lower bound of B.

The proof is incorrect

>>8987267
My proof or Rudin's, also I forgot to mention that L is bounded above by any element of B

>>8987267
>It's easy to find a subset of B not in L that contains the greatest lower bound of B.

what are you talking about? Greatest lower bound of B is a lower bound of B, therefore it's in L

>>8987251
>First, suppose S is an ordered field with the LUB property, take A, a nonempty subset of S bounded above and below. Consider L a subset of S consisting of the lower bounds of A, A is bounded below, so L is not empty, so it has a least upper bound, call it s.
Ok

>Now we must show s is a greatest lower bound of A. This is where I'm stuck, I don't know how to justify that s is in L
If you assume s is not in L, then there exists a point in S, say c, less than s. Now we must have that there exists another lower bound for S between s and c, otherwise c would be the real inf of S. But we cant have a lower bound of S greater than c. I'm not even gonna read what Rudin wrote, but its probably the same thing.

>Assuming this is true, though, can I say that for any x > s, there is some a in A such that a <x and then be done, as s is the glb of A?
Yes. It follows from the definition of the glb. For otherwise what you have listed as x would be the real glb.

>>8987267
>I don't know what I'm talking about but I'm just going to sperg out and say whatever the fuck I'm thinking

>>8987344
Ignore that bit about c being the real inf of S. There still must be another lower bound between c and s, otherwise the real inf would be less than OR equal to c.

>>8987344
Do you mean to say A whenever you say S?

>>8987251
>This is where I'm stuck, I don't know how to justify that s is in L.
What part of Rudin's justification you don't understand though? It seems pretty clear: any element less than s is not an upper bound of L (since s is the least upper bound) implying that any element less than s is not in A (since all A's elements are upper bounds of L) meaning that s ≤ any element of A - the definition of being a lower bound of A -> [math]s \in L[/math] since all lower bounds are.

>>8987383
So because all elements of A are upper bounds of L is logically equivalent to elements that are not upper bounds of L are not in A, ie elements less than the least upper bound are not elements of A, so elements of A are greater than or equal to s, so she is a lower bound of A, and is hence in L

>>8987421
correct