The Frattini subuniverse is the intersections of all proper maximal subuniverses

Question

From "Universal Algebra: Fundamentals and Selected Topics" of Clifford Bergman.

An element $a$ of an algebra $A$ is called a non-generator of $A$ if
for every $X \subseteq A$, $A = Sg(X \cup \{a\})$ implies $A = Sg(X)$.

(a) Prove that the set of nongenerators of $A$ forms a subuniverse,
$Frat(A)$ (called the Frattini subuniverse of $A$).

(b) Prove that $Frat(A)$ is the intersection of all maximal proper
subuniverses of $A$. (If you wish, you can assume that $A$ is
finite. To do the infinite case, you will need Zorn’s lemma.)

My solution for (a)

Suppose $a_1, …, a_n$ to be non-gerators of $A$ and $f$ a fundamental operation of A then:

$X \cup \{f(a_1,…, a_n)\} \subseteq X \cup \{a_1, …, a_n \} \cup \{f(a_1,…, a_n)\}$

It's easy to see that:

$Sg(X \cup \{a_1, …, a_n \} \cup \{f(a_1,…, a_n)\})=Sg(X \cup \{a_1, …, a_n \})=Sg(X)$

Hence $Sg(X \cup \{f(a_1,…, a_n)\}) \subseteq Sg(X)$ and the other verse of inclusion is trivial.

My question

I'm looking for a proof of (b).

Answer 1

Let 𝓜 = {Mᵢ : i ∈ I} be the collection of all maximal proper subuniverses of A.

Let Frat(A) be the collection of non-generators of A.

To show: ⋂ 𝓜 = Frat(A)

1. Frat(A) ⊆ ⋂ 𝓜. Let u be a non-generator of A and show that ∀i ∈ I, u ∈ Mᵢ as follows: Fix i. If u does not belong to Mᵢ, then Sg(Mᵢ,u) = A. But then Mᵢ = Sg(Mᵢ) = A, since u is a non-generator. On the other hand, Mᵢ is proper. This contradiction proves that u belongs to Mᵢ, and i was arbitrary.

2. ⋂ 𝓜 ⊆ Frat(A). Let u ∈ ⋂ 𝓜. We prove u is a non-generator. Suppose A = Sg (X ∪ {u}) for some X ⊆ A. To see that Sg(X) = A, suppose the contrary; i.e., Sg(X) ≠ A. Then (by Zorn's Lemma) there exists a maximal proper subuniverse M ≤ A containing Sg(X). But then u ∈ ⋂ 𝓜 ⊆ M implies that M contains both X and u and thus M contains A = Sg (X ∪ {u}), contradicting the fact that M is proper. ∎