Proof of theorem: Every proper ideal a is contained in some maximal ideal. I found a proof of this theorem in modern algebra 2.30

Proof: Set $S := \{ \text{ ideals } \mathfrak b \mid \mathfrak b ⊃ \mathfrak a \text{ and } \mathfrak b ∌ 1 \}$. Then $\mathfrak a ∈ S$, and $S$ is partially ordered by inclusion. Given a totally ordered subset $\{\mathfrak b_λ\}$ of $S$, set $\mathfrak b := \bigcup_\lambda \mathfrak b_λ$. Then $\mathfrak b$ is clearly an ideal, and $1 \notin \mathfrak b$; so $\mathfrak b$ is an upper bound of $\{\mathfrak b_λ\}$ in $S$. Hence by Zorn’s Lemma, $S$ has a maximal element, and it is the desired maximal ideal.

I'm not satisfied with the proof, with follow questions:


*

*Why $S$ is a set, $S$ may not be a set as Russell's paradox.

*How to apply Zorn’s Lemma, can I apply Zorn’s Lemma to the range of real number $(0,1)$, and say that range $(0,1)$ has a maximal element ?


I'm guessing that missing key to the proof is that to prove all partially ordered chain is finite.
Edit 1: still confused about something here
I'll start from an example:

Proof: $S = \{ 1 - \frac 1 n \mid n \in \mathbb N \}$ has a maximal element
Obviously $S$ is partially ordered, let $x = 1 - \Pi_{e \in S} (1 - e) = 1 - \Pi_{n \in \mathbb N} \frac 1 n$, then clearly $x \in S$; so $x$ is an upper bound of $S$. Hence by Zorn’s Lemma, $S$ has a maximal element.

I just copied the proof word by word, then got an obviously wrong result. Where does my proof go wrong ?
If I'm not allowed to construct $x$ this way then it is also not allowed to construct $\mathfrak b$ in the origin proof.
I think the truth is $x=1$ and $x \notin S$, and what about $\mathfrak 
 b = \bigcup_\lambda \mathfrak b_λ$ ? Things can go wild with infinite. I'm not satisfied with $1 \notin \mathfrak b$
 A: $S$ is a subset of the powerset of the elements in the ring. As such it can be easily constructed directly using the axioms of ZF set theory (the power set axiom and one of the axioms of comprehension / specification, specifically), starting with the set of all the elements of the ring (which by definition of a ring must be a set).
We cannot apply Zorn's lemma to $(0,1)$ with the standard ordering, because the hypothesis of Zorn's lemma isn't satisfied. Zorn's lemma says

If $(P, \geq)$ is a partial order such that every totally ordered subset of $P$ has an upper bound in $P$, then $P$ has a maximal element.

The interval $(0,1)$ with the standard ordering has itself as a totally ordered subset, and that subset doesn't have an upper bound in $(0,1)$. So Zorn's lemma cannot be applied.
For sets of ideals ordered by inclusion, on the other hand, a totally ordered subset is just a chain of ideals contained in one another. In that case, the union of those ideals is still an ideal, and the union of a totally ordered chain of ideals in $S$ is still an ideal in $S$. So any totally ordered subset of $S$ has an upper bound in $S$, and therefore $S$ must have a maximal element.
A: *

*$S$ is a set by the Axiom schema of specification.

*You cannot apply Zorn's lemma to the interval $(0,1)$ for the usual order  because not every chain in $(0,1)$ has an upperbound  in (0,1).

