Baby Rudin Theorem 1.11

  • #1

Main Question or Discussion Point

Theorem: Suppose S is an ordered set with the least-upper-bound property, B⊂S, B is not empty, and B is bounded below. Let L be the set of all lower bounds of B. Then α=supL exists in S, and α=infB.

Rudin proves that α=supL, α is an element of L and that α=infB.
For α to be sup i.e. lub it has to be in set of upper bounds. Here i quote rudin: "Since B is bounded below, L is not empty. Since L consists of exactly those y that are elemet of S which satisfy the inequality y≤x for every x from B, we see that evry x that is in B is an upper bound of L. Thus L is bounded above. Our hypothesis about S implies therefore that L has a supremum in S; call it α. If γ<α then γ is not an upper bound of L, hence γ is not an element of B. It follows that α≤x for every x in B. Thus α is an element of L. If α<β then β is not an element of of L, since α is an upper bound of L. We have shoh that α is an element of L but β is not element of L if α<β. In other words , α is lower bound of B, but β is not is not if α<β. This means that α=infB."


So since b consists of all of upper bounds of L and α is least upper bound, can i say that α is also an elemet of B?

If that is the case can i then say that maxL=supL=infB=minB?
Because α is an element of L and at the same time α=supL then maxL=α. If it were not the cas that maxL=α then there would be some other element β in L, that is greater then α. But this would contadict the fact that supL=α. Same goes for minB=α just reveresed.

Are these conclusions correct?

P.S. Why i don't have LATEX reference in editor?
 
Last edited:

Answers and Replies

  • #2
verty
Homework Helper
2,164
198
Something is very wrong. Let S = (0,1) ##\cup## (2,3), B = (2,3). Now B is bounded below, eg by 0 and 1/2, but sup(L) = 2 ##\not\in## S.
 
  • #3
Something is very wrong. Let S = (0,1) ##\cup## (2,3), B = (2,3). Now B is bounded below, eg by 0 and 1/2, but sup(L) = 2 ##\not\in## S.
If i may usk where does least-upper-bound property come into play in your counter example?
 
Last edited:
  • #4
Something is very wrong. Let S = (0,1) ##\cup## (2,3), B = (2,3). Now B is bounded below, eg by 0 and 1/2, but sup(L) = 2 ##\not\in## S.
Since you gave counter example you assumed that S is an order set with least-upper-bound property. (Right?) If so then this means that if E is subest of S, and E is non empty and E is bounded above , then sup E exists in S.
So let S = (0,1) ##\cup## (2,3) and Let E= (2,3). E is subset of S and it is no empty and it is bounded above and more importantly supE should exists in S. But E=(2,3) fails to have sup in S. Sup of E is 3 and it is not in S, because i can come infinetly close to 3 but i will never find element that is greater then all other elements in E. Therefore S fails to be an order set with least-upper-bound property and there fore your contuer example is false. Is my reasoning correct here?
 
  • #5
verty
Homework Helper
2,164
198
Since you gave counter example you assumed that S is an order set with least-upper-bound property. (Right?) If so then this means that if E is subest of S, and E is non empty and E is bounded above , then sup E exists in S.
So let S = (0,1) ##\cup## (2,3) and Let E= (2,3). E is subset of S and it is no empty and it is bounded above and more importantly supE should exists in S. But E=(2,3) fails to have sup in S. Sup of E is 3 and it is not in S, because i can come infinetly close to 3 but i will never find element that is greater then all other elements in E. Therefore S fails to be an order set with least-upper-bound property and there fore your contuer example is false. Is my reasoning correct here?
I wasn't sure what the "least upper bound property" was but I couldn't see what would make the theorem true. This time let S = (0,1] ##\cup## (2,3], B = (2,3].

Oh, I realize now what is going on. It has be given implicitly that only points of S are lower bounds. With this premise, the theorem is true.

Back to your original question, no, my counterexample: S = (0,2], B = (1,2).
 
  • #6
26
0
You should definitely look at Real Mathematical Analysis by Charles Chapman Pugh. I find it much easier to read than Rudin. I honestly don't know why Baby Rudin is preferred by so many, I feel like Baby is sort of outdated and lacks organization of thoughts in some of his proofs.

Your conclusion, however, is false. We say that a is a maximal (or minimal) element of A if a is contained in A and a=sup(A) (or a=inf(A)). I looked at the proof in Baby Rudin, and it didn't explicitly say that alpha is contained in B nor do I see how it could be inferred. Of course there are cases in which the intersection of two closed sets is nonempty and in this case, if the intersection was a singleton, then yes sup(L)=max(L)=inf(B)=min(B). Looking ahead though, to get you ready for Dedekind cuts and cut arithmetic, I think that the point of the theorem is actually to show that a supremum or infimum of a set need not be a member of the set. And this is always the case when the two subsets are complements in the parent set S. The assumption is "Let L be the set containing all lower bounds of B". This is why sup(L) is contained in L and so sup(L)=max(L)=inf(B).
 

Related Threads on Baby Rudin Theorem 1.11

Replies
3
Views
2K
  • Last Post
Replies
5
Views
4K
Replies
7
Views
3K
Replies
3
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
11
Views
2K
Top