The chain of subfields continues indefinitely, each set strictly containing the previous. Here's the breakdown:

Main Question: Infinite Chain?

Yes, the chain ( A_0 \subsetneq A_1 \subsetneq A_2 \subsetneq \ldots ) continues forever.

• Base Case: ( A_0 = \mathbb{Q} ). ( A_1 = S_{\mathbb{Q}} ) includes all real algebraic numbers, which strictly contains ( \mathbb{Q} ) (e.g., ( \s...

First, let me fix a small point about interpretation. For real exponents, x↦x^e is naturally defined on (0,∞) by x^e = exp(e ln x). For negative x the expression x^e is real only for restricted (typically rational) exponents. Nothing below hinges on the behavior at negative x, so you may safely read the statements as being made for positive real solutions.

1. SQ is exactly the set of (real) algeb...

Throughout let
 S­­₍F₎ = { x ∈ ℝ : 0 = a₁x^{e₁}+⋯+aₙx^{eₙ} for some n≥1 with pair-wise different exponents eᵢ∈F and non–zero coefficients aᵢ∈F }.
Put A₀ = ℚ and A_{n+1}=S(A_{n}).
The question is whether the inclusions A₀⊊A₁⊊A₂⊊… stay proper for every finite n, whether every A_{n} is a field, and whether numbers such as e ever lie in one of the A_{n}.

1. The first two steps
    A₁ is exac...

Here is what is known (and not known) about your tower

1. Definitions. Let F⊆ℝ be any set and define S_F = { x>0 : there is a “generalized‐polynomial” 0 = ∑{i=1}^n a_i x^{e_i}, with a_i∈F{0}, e_i∈F distinct } In particular A_0 = ℚ, A{n+1} = S_{A_n}.

2. S_F is a subfield (of the positive reals) as soon as F is a field.
   (i) M...

<think> We are given a chain of sets defined as follows: - \( A_0 = \mathbb{Q} \) - \( A_{n+1} = S_{A_n} \), where for any subset \( F \subset \mathbb{R} \), \( S_F \) is the set of real numbers that are roots of a "generalized polynomial" with exponents in \( F \) and coefficients in \( F \setminus \{0\} \).

Specifically, ( S_F = \left{ x \in \mathbb{R} : \sum_{i=1}^n a_i x^{e_i} = 0 \text{ ...

We write
 S F = { x∈ℝ : ∑ i a i x e i = 0 for some n, a i ,e i ∈F,   e i mutually different }.
Put A 0 =ℚ and A n+1 =S A n .

1. The chain A 0 ⊂A 1 ⊂A 2 ⊂⋯ is strictly increasing.

• A 0 ⊂A 1 because √2 is the unique positive solution of x 2 −2=0, so √2∈A 1 \A 0 .

• Fix n≥1 and choose an irrational algebraic number β∈A n (for instance β=√2, which is already in A 1 ⊂A n ).
C...

Constructing an Infinite Chain of Subfields of 'Hyper' Algebraic Numbers

This question explores a fascinating recursive construction of sets that generalizes the concept of algebraic numbers.

Understanding the Setup

We start with a set and define as:

An excellent and insightful question that touches upon the frontiers of transcendental number theory. Let's break down the components of your question.

Summary of the Answer

1. Is A_i always a field? No. A_0 = \mathbb{Q} and A_1 = \mathbb{A}_\mathbb{R} (the real algebraic numbers) are fields. However, for n \ge 2, A_n is not known to be a field, and it is not expected to be one...

This is a fascinating question about iterating a generalization of algebraic closure. Let me address your questions systematically.

Main Question: Does the chain continue forever?

Yes, the chain continues forever. Here's why:

For each , we can construct an element in explicitly. Consider the equation: where is any element of ...