# Clarification needed with Rudin Theorem 1.11

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!

• You seem to think of $L$ and $B$ as disjoint. – Phira Jun 19 '18 at 22:39
• 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 $0\in L$ – Doug M Jun 19 '18 at 22:43
• This has nothing to do with set-theory, so I've removed the tag. Please do not add it back, and read the description of the tag to clarify its scope. – Andrés E. Caicedo Jun 22 '18 at 11:50

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 $B$.

• Could you explain more as to how it matters not whether it is in B? – Red Jun 19 '18 at 22:57
• An infimum does not need to be in the set but it is allowed to be. If it is inside, it is also a minimum, but it is still also an infimum. – Phira Jun 19 '18 at 23:01

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 $B.$

• I see thank you that makes seems to make sense. It is by the very definiton of L that we can say $\alpha$ is an element of it. – Red Jun 19 '18 at 23:31
• Does my attempt at explaining the lines previous to this make sense or am I perhaps not writing clear enough? – Red Jun 19 '18 at 23:32

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 $B$.

I will try to answer my own question since I wasn't completely satisfied with answered provided although they did help.

"If $\gamma \lt \alpha$ then $\gamma$ is not an upper bound of L, hence $\gamma \notin B$."

So I was having extreme trouble coming to terms with the fact that why would Rudin need to write this into his proof. And the reasoning I gave in my question was wrong but I don't know whether I should leave it there(because it was an honest attempt) so I trust someone in the community will edit the question if they see it wise to do so.

So far in the proof we have shown that the $sup\ L= \alpha$ but we have said nothing about the elements such as $\gamma$ that are less than $\alpha$. We use these "nearby" elements to talk about other mathematical objects** of importance in this case $\alpha$. We use, by definition(definition of an least upper bound in Rudin's Book), the fact that $\gamma$ is not an upper bound of $L$, and since every element that is in $B$ is an upper bound of $L$ it follows $\gamma \notin B$ either.

"It follows that $\alpha \le x$ for every $x \in B$. Thus $\alpha\in L.$"

Once again since every element $x \in B$ is an upper bound of $L$ it follows from continuing from the argument above that $\alpha \le x$. And by definition of being a lower bound we can say of alpha that $\alpha \in L.$

And ofcourse later on in the proof we go on to show that $\alpha \in B$ as well as couple of the other answers previously touched on.

So in conclusion Rudin included this portion of the proof and wrote it this way to show that the existence of elements $\gamma$ of $L$ help prove $\alpha \in L$ which is intuitively obvious but still needs to be stated.

**I am told this is a common technique in analysis and is similar to the epsilon-delta proofs or neighborhood proofs which seems to make some sense to me. I'd appreciate a comment on this.

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

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.