1 seems intuitive to me, but it's not the supremeum of $E,$ not according to Rudin's book, 1 can't be an upper bound because 4 exists and 4>1. Maybe the problem is that it fails to specify what $\gamma$ is.

@StackQuest (ii) in Definition 1.8 says that "if $\gamma < \alpha$ then $\gamma$ is not an upper bound". Here $\alpha$ is $1$, and $4 \not< 1$.

But $\alpha$ is 4 because $1<4.$

@StackQuest therefore $4$ is an upper bound but not the least upper bound, and $1$ is the least upper bound.

But $4$ isn't in $E$. If $4$ were in $E$ then $1$ would be an upper bound for $E$. In fact of course $$ \forall\,x\in E\ x\le 1.$$ This is exactly the statement that $1$ is an upper bound for $E$.

What part of the definitions says $4$ needs to be in $E$?

@JamieRadcliffe I think you mean "...then $1$ would not be an upper bound for $E$".

@StackQuest see my edit.

Here's what I read according to (ii) of definition 1.8: $1$ and $4$ are both candicates for being upper bounds of $E.$ But, $1$ CANNOT be an upper bound because $4>1$ (this is a direct consequence of (ii)). Let $\gamma = 1$ and $\alpha = 4,$ then it is in fact true $1<4.$ Can you specify what $\gamma$ is an element of? Rudin's book appears to make no mention of this.

You mixed up the role of $\alpha$ and $\gamma$ in the definition. To show that $1$ is the least upper bound, we want to show that: 1. $1$ is an upper bound of $E$. 2. If $\gamma < 1$ then $\gamma$ is not an upper bound of $E$. So $\gamma = 4$ does not violate condition (2).

Okay, both $1$ and $4$ possess the quality that $1 \geq x$ for all $x \in E,$ and $4 \geq x$ for all $x \in E.$ Do you agree with that?

Yes, both $1$ and $4$ satisfy criterion (1).

Also $\gamma$ can be any element in $S$.

Okay, then first just looking at criterion (1), both $1$ and $4$ are upper bounds of $E,$ correct?

Yes, that's correct.

Okay, but $\gamma = 1$ is less than $4,$ so $1$ now suddenly out of nowhere, $1$ spontaneously stops being an upper bound of $E,$ that's the first part that makes no sense. Well, $4$ is greater than $1$ so $1$ cannot be an upper bound. That follows from criterion (ii). All my intuition and several other people's intuition says $1$ is the least upper bound, but according to Rudin, $1$ is not even an upper bound at all.

Firstly, violating (2) does not make it not an upper bound, but make it not a least upper bound. Secondly, as I said in the previous comments, to check that $1$ is the least upper bound, you need to check that 1. $1$ is an upper bound of $E$ and 2. If $\gamma < 1$ then $\gamma$ is not an upper bound of $E$. You're letting $\gamma = 1$ and $\alpha = 4$, but that's irrelevant in the context of checking if $1$ is the least upper bound.

Sorry to be crude in this way, but why should I believe you over Rudin? Rudin says if $\gamma < \alpha$ then $\gamma$ is not an upper bound. Well, you're out of luck, $1 < 4$ therefore $1$ is not an upper bound, this is straight from the definition. Otherwise, it may be a typo.

I'm not sure what kind of baby Rudin you're reading, but Definition 1.8 of Rudin clearly says least upper bound. For a sanity check, is your version of baby Rudin the same as this version?

I'm looking at the same book and edition as you. Rudin clearly says "If $\gamma < \alpha$ then $\gamma$ is not an upper bound of $E.$" So $1$ is not an upper bound of $E$ because $4$ is an upper bound and $1 < 4.$

You're saying that $4$ satisfies criterion (2) of Definition 1.8. That's false because $4$ is an upper bound not a least upper bound. Not all upper bounds satisfy (1) and (2) Definition 1.8, only least upper bounds.

$4$ exactly satisfies criterion (ii). $4 = \alpha.$ $1$ however, does not because $4 > 1.$ I know this is confusing, but Rudin must be correct, $\gamma = 1$ is $< 4,$ so $1$ cannot be an upper bound of $E.$ This is straight from the definition.

Can you elabroate how does $4$ satisfy criterion (2)? Cos I'm certain it doesn't.

$4$ is an upper bound of $E,$ we both agreed on that. All criterion (ii) does, by definition, is simply specify what numbers are not upper bounds. $1$ is less than $4,$ so $1$, by criterion (ii), cannot be an upper bound of $E.$

That's not the purpose of criterion (2). Criterion (2) places a requirement for an upper bound to be a least upper bound. So we can have upper bounds who fail to satisfy criterion (2).

I can't say what any alleged philosophical "purpose" is. Rudin says if $\gamma < \alpha$ then $\gamma$ is not an upper bound of $E.$ Well, $4$ is an upper bound, that's our $\alpha,$ and $1 < 4,$ so clearly $1$ cannot be an upper bound of $E.$ This follows from criterion (ii).

There's no philosophical aspect in my statement. I'm saying that your interpretation of criterion (2) (or just Definition 1.8 overall) is wrong. Rudin says that if $\alpha$ is a least upper bound, and $\gamma < \alpha$, then $\gamma$ is not an upper bound. But since $4$ is not a least upper bound, it doesn't imply that $1$ is not an upper bound.

Rudin says no such thing, Rudin says $\alpha$ is an upper bound of $E.$ The number $4$ fits this criterion. Now, criterion (ii) says "If $\gamma < \alpha$ then $\gamma$ is not an upper bound of $E.$" We're out of luck, sorry, by this definition, $1$ cannot be an upper bound because $4 > 1.$ If you hope to resolve this, throw away all your preconceived notions of whatever you think this "should" mean, I don't share those preconceptions, all I have is Rudin's definition, and then perhaps other definitions like wikipedia's.

You seem to think that if $\alpha$ satisfies criterion (1), then it must satisfy criterion (2). That's neither true nor what Definition 1.8 is saying. Definition 1.8. lists two completely different criterions for $\alpha$ to be a least upper bound - (1) $\alpha$ is an upper bound, and (2) If $\gamma < \alpha$ then $\gamma$ is not an upper bound. Nobody says anything about (1) $\implies$ (2) or (2) $\implies$ (1).

I agree that we are not given that (i) $\implies $ (ii) nor vice versa. My interpretation still stands. By criterion (ii), $1$ cannot be an upper bound because $4$ is an upper bound and $4 > 1.$

I'm going to make this one comment and then be done. It has been established that $1,2,3,4$ are all upper bounds for your set. According to Rudin's definition, $4$ cannot be the least upper bound because there are upper bounds that are less than $4,$ namely $1,2,3.$ For a similar reason neither $3$ nor $2$ can be upper bounds. Of all possibilites, only $1$ satisfies the properties that it is an upper bound and that no smaller number is an upper bound. Walk away from this for an hour or so, come back with a refreshed mind, and think carefully about what the two definitions are actually saying.

I agree with Chris that it's probably unproductive at this moment to elaborate any further, and you should take a break from this for now.

Rudin clearly implies numbers less than $4$ cannot be upper bounds of $E.$ This follows directly from criterion (ii). Prove me wrong please. As I advised, do not rely on what you "think" the definition "should" be saying if you are to understand the dilemma here. One way to resolve this is to say Rudin's definition is garbage and look at another like Wikipedia's.

There's even a typo on the same page, it says "Since A has no lasgast member..." I think none of you understood how the logic followed from Rudin's definition, because it doesn't, it is incomplete or has a typo. Instead, much more reasonably and accurately, you constructed your own understandings from your professor's lectures, discussions and homework problems, and then tried to convince yourselves that Rudin's definition specified your clearly new and clearly constructed understanding of your own gaps that you filled in. Rudin doesn't even specify what $\gamma$ is an element of, c'mon.