0

Rudin's theorem 1.11 states:

Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $B$ is not empty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$. Then $\alpha = \sup L$ exists in $S$, and $\alpha = \inf B$. In particular, $\inf B$ exists in $S$.

I am having trouble understanding why $L \subset S$. This has to be the case so that we can invoke the LUB property to conclude that $\sup L$ exists in $S$. After a lot of searching, the answer seems to be that "$S$ is the universe; nothing exists outside of $S$." I'm struggling to understand why that follows from the problem. What if, for example, $S$ is the set $\mathbb{Z}$, $B$ is the positive integers, and $L$ is the set of all lower bounds in $\mathbb{R}$? That is, $L = (-\infty, 0)$. Surely, $L \not \in S$.

If we went a step further and defined $L$ as the set of all lower bounds of $B$ in $S$, the proof would make much more sense to me. Is that what Rudin, implicitly, means?

John P.
  • 2,136
  • A lower bound for $B$ is an element $x$ of $S$ such that $x \leq y$ for all $y \in B$. In your example $L$ is the set of all non-positive integers. – Kavi Rama Murthy Jul 01 '20 at 08:49
  • But why can't $S$ be a subset of another ordered set, $T$? In that case, a lower bound for $B$ may be an element of $T$. – John P. Jul 01 '20 at 08:53
  • 1
    Your argument is something lime this: If a book is discussing real numbers and there is a result which says $x^{2} \geq 0$ for all $x$ you are giving $i^{2}=-1$ as a counter-example. You have to pay attention to the context. – Kavi Rama Murthy Jul 01 '20 at 08:58
  • Is it fair to say then that $L$ is defined as the set of lower bounds of $B$ that are in $S$? If this is true because of the context of the problem, I'm fine with the result. – John P. Jul 01 '20 at 09:11
  • 1
    Yes, exactly that's the correct interpretation. – Berci Jul 01 '20 at 09:14
  • 1
    Yes, that is exactly what $L$ is, in this context. – Kavi Rama Murthy Jul 01 '20 at 09:14

2 Answers2

2

Simply the sentence 'Let $L$ be the set of all lower bounds of $B$.' should be implicitly understood as

Let $L$ be the set of all lower bounds $s\in S$ of $B$.

Berci
  • 90,745
0

I struggled with this myself, and I was looking for a bit more rigorous explanation. Here's what I thought.

To answer this question more explicitly without relying on any "intuition" or contextual meanings, observe the following.

Recall 1.7 Definition in the textbook (changed to bounded below, as suggested in the textbook): Suppose S is an ordered set, and $E\subset S$. If there exists a $\beta \in S$ such that $x \ge \beta$ for every $x \in E$, we say that $E$ is bounded below, and call $\beta$ a lower bound of $E$.

(I am not quoting above as I have changed the original text of the textbook to suit the definition we are needing)

We are going to transform this into two logical pieces.

Let $p = $ "There exists a $\beta \in S$ such that for every $x \in E$, $x \ge \beta$"

Let $q = $ "$\beta$ is a lower bound of $E$"

Therefore $p \implies q$. But $q \implies p$, therefore $p \Leftrightarrow q$

In other words, If $\beta$ is a lower bound of $E$, then $\beta \in S$ such that for every $x \in E$, $x \ge \beta$.

Meaning that it is in the definition of a "lower bound" to exist in the $E$'s superset, which is $S$ in this case. And same can be said about the Theorem. By definition, the lower bounds of $B$ exist in it's superset which is $S$.

Shayan
  • 65