# Rudin's Principles Theorem 1.11 (supremum, infimum)

## Main Question or Discussion Point

Mod note: Edited by removing [ sup ] tags.
To the OP: Please don't fiddle with font tags, especially the SUP tag, which renders what you write in very small text (superscript).

Hello everyone

I have just started studying mathematics at university this summer and I have decided to supplement my scheduled classes and accompanied books by working through Rudin's analysis book for undergraduates. I am not completely certain, though, that I understand theorem 1.11 found on page five and I hope for some of you to help me a little.

As I understand, a set, say S, has the least-upper-bound property if given any non empty subset of S, say E, which is also bounded above its supremum must lie in S.
So since T={x is in Q: 2<x^2} is a subset of Q and the inf(T)=sqrt(2) is not in Q itself Q does not have the least-upper-bound property, right?

So in my own words theorem 1.11 says something like:

Let S be any set which has the least-upper-bound property and let B be a non empty subset of S which is bounded below. Since it is bounded below there must exists another set whose elements are all lower bounds of B and which consists and all possible lower bounds of B; although the set could theoretically consist of just one element. Call this set L. Conversely, every element of B must function as a upper bound of L. Because S has the least-upper-bound property every subset of S which is not empty and has an upper bound must also have an supremum, I.e. sup(L) exists. Call it £. But because B is defined as consisting of all upper bounds of L we know that £ must lie in B. Also, for every x>£, x must be in B and B only. If we negate it we get x<£ or x=£ and equivalently the opposite of x is only in B, namely x could be outside B. But since B is arbitrary the only way this be certain is for x to be in L. Hence £ is in L. But because every lower bound of B is in L, it's greatest lower bound must also be in L and the largest element in L is £ so it follows sup(L)=inf(B).

I am truly sorry to use that much space, but I cannot formulate myself any better at this point.

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Stephen Tashi
So in my own words theorem 1.11 says something like:
If you are asking for an evaluation of your own words, if would help if you quoted the wording of the theorem from the book.

HallsofIvy
Homework Helper
Hello everyone

I have just started studying mathematics at university this summer and I have decided to supplement my scheduled classes and accompanied books by working through Rudin's analysis book for undergraduates. I am not completely certain, though, that I understand theorem 1.11 found on page five and I hope for some of you to help me a little.

As I understand, a set, say S, has the least-upper-bound property if given any non empty subset of S, say E, which is also bounded above its supremum must lie in S.
So since T={x is in Q: 2<x^2} is a subset of Q and the inf(T)=sqrt(2) is not in Q itself Q does not have the least-upper-bound property, right?
You seem to be confusing "upper bound" and "lower bound". Since sqrt(2) is a lower bound for T, it has nothing to do with upper bounds on subsets of T.

So in my own words theorem 1.11 says something like:

Let S be any set which has the least-upper-bound property and let B be a non empty subset of S which is bounded below. Since it is bounded below there must exists another set whose elements are all lower bounds of B and which consists and all possible lower bounds of B; although the set could theoretically consist of just one element. Call this set L. Conversely, every element of B must function as a upper bound of L. Because S has the least-upper-bound property every subset of S which is not empty and has an upper bound must also have an supremum, I.e. sup(L) exists. Call it £. But because B is defined as consisting of all upper bounds of L we know that £ must lie in B. Also, for every x>£, x must be in B and B only. If we negate it we get x<£ or x=£ and equivalently the opposite of x is only in B, namely x could be outside B. But since B is arbitrary the only way this be certain is for x to be in L. Hence £ is in L. But because every lower bound of B is in L, it's greatest lower bound must also be in L and the largest element in L is £ so it follows sup(L)=inf(B).

I am truly sorry to use that much space, but I cannot formulate myself any better at this point.