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I am reading Rudin's Principles of Mathematical Analysis and I have some questions regarding the proof of Theorem 1.11.

Theorem 1.11. Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $S$ 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 $\alpha = \inf B$.

Question 1: In the proof above, I do not understand the reasoning behind "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 $\alpha\in L.$" In other words I interpret this as saying that if there was an element less than the supremum of $L$ then it couldn't be an upper bound of $L$ nor could it be an element of $B$, which seems like an obvious thing to say why state this?

Question 2: So it goes on to say that $\alpha$ is less than or equal to every element of B but how does this show $\alpha \in L$?

Question 3: Also can someone explain how the sentence "In other words, $\alpha$ is the lower bound of $B$ but $\beta$ is not if $\beta\gt \alpha.$" implies "$\alpha=inf B$"

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    $\begingroup$ To your "Question 1" I'd only say that this is an introductory textbook; it's likely one will encounter things that seem "obvious." Next $L$ is precisely the set of lower bounds; $\alpha$ satisfies the definition of a lower bound. Finally, careful with wording: $\alpha$ is ${\bf a}$ lower bound of $B$, not ${\bf the}$. No larger $\beta$ is a lower bound of $B$, so by definition $\alpha$ is the greatest lower bound of $B$. $\endgroup$ – John Brevik Jun 19 '18 at 13:58
  • $\begingroup$ The point is that the 3 sentences "$S$ has both the LUB property and the GLB property ", ..."$S$ has the LUB property" ,... "$S$ has the GLB property" are all equivalent to each other. $\endgroup$ – DanielWainfleet Jun 19 '18 at 14:35
  • $\begingroup$ @JohnBrevik . Good point about careful wording. $\endgroup$ – DanielWainfleet Jun 19 '18 at 14:38
  • $\begingroup$ @JohnBrevik When it is proved there is a supremum of L in S for all we know that supremum could be in B. So could the reason for this obvious statement be to say: that if $\gamma \lt \alpha$ then $\gamma$ obviously can't be in B but also $ \alpha $ is can only exist as a number less than or equal to all numbers in B.But I don't understand how this means $\alpha \in L$?? Because $\alpha $ could be an element of B couldn't it? $\endgroup$ – Red Jun 19 '18 at 17:05
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    $\begingroup$ @Red Note that $\alpha$ can be an element of $B$ as well as of $L$. For example, take $S = \mathbb{R}$ and $B = [0,1]$. Then $L = (-\infty,0]$, and $\alpha = 0 \in L \cap B$. So, $\alpha$ being an element of $B$ does not necessarily contradict the statement that $\alpha$ lies in $L$. $\endgroup$ – Brahadeesh Jun 20 '18 at 3:33
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In other words I interpret this as saying that if there was an element less than the supremum of $L$ then it couldn't be an upper bound of $L$ nor could it be an element of $B$, which seems like an obvious thing to say why state this?

Rudin is spelling every reason and conclusion in detail because this theorem is in an introductory chapter. Indeed, it is worthwhile as a beginner to try and pin down why this "obvious" fact is true.

So it goes on to say that $\alpha$ is less than or equal to every element of $B$ but how does this show $\alpha \in L$?

This is simply because $L$ is defined to be the set of all elements that are less than or equal to every element of $B$. If $\alpha$ satisfies this condition, then $\alpha$ must be in $L$ by the definition of $L$.

Also can someone explain how the sentence "In other words, $\alpha$ is the lower bound of $B$ but $\beta$ is not if $\beta > \alpha$." implies "$\alpha = inf B$"

The infimum of a set that is bounded below is defined to be that real number which is a lower bound, and is greater than any other lower bound. Rudin says in the former sentence that $\alpha$ is a lower bound of $B$, and that no number larger than $\alpha$ is a lower bound, that is, every other lower bound must be less than or equal to $\alpha$. Hence, $\alpha = \inf B$ by the definition of infimum.


Here is my elaboration of the proof given by Rudin. Do read and see if it throws more light on each of the steps in the proof.

The goal is to show that $\sup L =: \alpha = \inf B$. So, we want to show that $\alpha$ satisfies:

  1. $\alpha \leq x$ for all $x \in B$; and
  2. If $\beta \leq x$ for all $x \in B$, then $\beta \leq \alpha$.

That is, we want to show that

  1. $\alpha \in L$; and
  2. $\beta \in L \implies \beta \leq \alpha$.

For the first part, if $\alpha \not \in L$, then $\alpha$ is not a lower bound for $B$, so there exists $x \in B$ such that $x < \alpha$. But, from the definition of supremum, there will exist an element $\gamma \in L$ such that $x < \gamma < \alpha$. But this is a contradiction, because $\gamma$ is a lower bound for $B$, so $\gamma \not > x$. Hence, $\alpha \in L$.

For the second part, let $\beta \in L$. If $\beta > \alpha$, then from the definition of supremum, there will exist an element $x \in B$ such that $\beta > x > \alpha$. But this is a contradiction, because $\beta$ is a lower bound for $B$, so $\beta \not > x$. Hence, $\beta \leq \alpha$.

$$\tag*{$\blacksquare$}$$

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You're right interpreting the first statement like this. But the author writing is more rigorous and I guess that's expected, as the excerpt is from the first chapter, where more attention is paid to details.

By definition we have that $L=\{y \in S \mid y \le x; \forall x \in B\}$, hence as $\alpha \le x; \forall x \in B$ we have $\alpha \in L$.

For the last part we have that $\alpha \le x; \forall x \in B$. To prove that $\alpha = \inf B$ we need to prove that if $\beta$ is a lower bound of $B$, then $\beta \le \alpha$. Indeed, assume $\beta > \alpha$, then as stated we have that $\beta$ isn't a lower bound of $B$, which is a contradiction. Hence the proof.

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