Clarification needed with Rudin Theorem 1.11

Question

I have asked this question before but I am rephrasing it here ( and deleting the previous post) as I slightly more nuanced take on the question. In Rudin Principles of Mathematical Analysis we have:

Theorem 1.11. 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 $S$.

Proof. Since $B$ is bounded below, $L$ is not empty. Since $L$ consists of exactly those $y \in S$ which satisfy the inequality $y \leq x$ for every $x \in B$, we see that every $x \in B$ 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 $\alpha$.

If $\gamma < \alpha$ then (see Definition 1.8) $\gamma$ is not an upper bound of $L$, hence $\gamma \not\in B$. It follows that $\alpha \leq x$ for every $x \in B$. Thus, $\alpha \in L$.

If $\alpha < \beta$ then $\beta \not\in L$, since $\alpha$ is an upper bound of $L$.

We have shown that $\alpha \in L$ but $\beta \not\in L$ if $\beta > \alpha$. In other words, $\alpha$ is a lower bound of $B$, but $\beta$ is not if $\beta > \alpha$. This means that $\alpha = \inf B$.

I understand the following portion of the proof as follows below:

"If $\gamma \lt \alpha$ then $\gamma$ is not an upper bound of L, hence $\gamma \notin B$. It follows that $\alpha \le x$ for every $x \in B$. Thus $\alpha\in L.$"

We have proven the there is a supremum of $L$ somewhere in $S$ but for all we know this supremum, which we call $\alpha$, could be in $B$. So to prove this point Rudin uses the definition of the least upper bound to remind us that $\gamma$ cannot be in $B$ because $\gamma$ must always be less than the supremum $\alpha$. But note also that since $\alpha$ must be a lower bound of $B$ so it follows from this that $\alpha \le x$ for every $x \in B$.

In other words Rudin uses the line above to cement the place of $\alpha $ as being between $L$ and $B$.

I have two questions dear reader:

1. After staring at this question as an undergrad novice for 12 hours is my explanation above some what close to correct?

2. Rudin ends this paragraph obliquely :"Thus $\alpha\in L.$"

How is this the case? Could it not be that $\alpha$ be in $B$ if in the previous line $\alpha = x$ for some $x \in B$. I am asking how does this follow from line reasoning previous in the paragraph? I am really passionate about understanding this proof so any effort you put forth will be greatly appreciated. Thanks in advance!

Answer 1

Your question 2: We have just seen that $\alpha$ is a lower bound of $B$, so by definition of $L$, it is in $L$. We neither know nor care whether it is in $B$.

Your question 1: It matters not whether it is in $B$.

My summary: I only have sups and would like to prove that I have infs, too,so I define inf as the sup of all lower bounds of $B$ and check the two property of an infimum: lower bound and bigger numbers are not lower bounds. In the first part we really only want to check that it is smaller than or equal to each element of $B$.

Answer 2

It is possible that $\alpha$ is in both $L$ and $B.$ For example let $B$ be the interval $[0,1].$ Then $0 = \inf B$ and $0\in B$ and $0\in L.$

The proof says that $L$ is bounded above. (all $x \in B$ are upper bounds of $L$). And since $L$ is a subset of $S$ and $S$ has a leas upper bound property. $L$ has a least upper bound.

Everything smaller than $\alpha$ is not in $B.$ $\alpha$ is a lower bound for $B.$ Since $L$ is the set of all lower bounds of $B,\alpha$ is in $L.$

And then he shows that $\alpha$ is not just a lower bound of $B$ but is, in fact, the greatest lower bound of $B.$

Answer 3

The OP states

I am really passionate about understanding this proof ...

To understand an abstract proof, a student can think about simpler concrete examples. It turns out that $\mathbb Z$ has the lub property (in fact, every set bounded above has a greatest element).

Here is a fun 'warm-up' exercise:

Exercise: Let $S$ be an ordered set where every nonempty subset bounded above has a greatest element. Show that any nonempty subset bounded below has an smallest element. Also, if $s \in S$ is not the greatest element of $S$, then we can 'go up 1 step' to the next greater element. Similarly, if $s \in S$ is not the smallest element of $S$, then we can 'go down 1 step' to the next smaller element.

To prove Rudin's theorem you must 'finesse' the logic employed in working on this exercise, being able to 'gently' handle '$\le$ to $\lt$ and $\ge$ to $\gt$' transitions.

How about cooking up an example for Rudin's theroem. If $B$ has a smallest element there is not much work to do. Suppose

$\quad B = \{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots, \frac{1}{n},\dots\}$

$B$ is bounded below and the set of all lower bounds $L$ is the set of non-positive numbers $L$. The $\text{sup(}L\text{)}$ is equal to $0$ which is also equal to the infimum of the set $B$.

Answer 4

This is not an answer, but too long a comment to fit in comment box.

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The real issue here is to prove that $\alpha=\inf B$. It is best to prove this using definition of infimum. First we prove $\alpha\leq x$ for all $x\in B$. To do so let us assume on the contrary that there is an $x\in B$ with $\alpha >x$. Now $\alpha=\sup L$ and hence there is a $y\in L$ such that $x<y\leq \alpha$ and this is absurd as $y$ is a lower bound for $B$ and hence $y\leq x$. This contradiction proves that $\alpha\leq x, \forall x\in B$.

Next note that if $\beta>\alpha=\sup L$ then $\beta\notin L$ and hence there is an $x\in B$ with $x<\beta$. And it is now clear that $\alpha=\inf B$.

Rudin's focus on showing that every member of $B$ is an upper bound for $L$ is entirely unnecessary. It's best to stick to the usual definition of $L$ as the set of lower bounds of $B$.

Also note that since $\alpha\leq x, \forall x\in B$ it follows that $\alpha\in L$. But the fact that $\alpha$ lies in $L$ or in $B$ is entirely irrelevant to the fact that $\alpha=\inf B$ and thus highlighting this fact is a mere distraction.