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 exactly the field of algebraic real numbers ℚ̄.

 Proof. If the exponents that occur are rational, write them with common denominator m. Putting y=x^{1/m} turns the given equation into an ordinary polynomial over ℚ, so x is algebraic. Conversely every algebraic number satisfies a usual polynomial, hence belongs to S(ℚ).

 A₂ is strictly larger than A₁.
Indeed √2∈A₁ and the equation x^{√2}-2=0 (coefficients 2, –1, exponent √2) shows 2^{1/√2}∈A₂. By the Gelfond–Schneider theorem 2^{1/√2} is transcendental, so A₁⊊A₂.

2. Countability
   If F is countable then the collection of “generalised polynomials’’ that may be used is also countable, every such equation has only finitely many real roots, and therefore S(F) is again countable. Hence every A_{n} is countable. Thus the chain can never exhaust ℝ, but it might conceivably stabilise at some countable stage.

3. Does the chain certainly go on for ever? (main question)
   At present this is not known. What is needed is, given a countable field F⊆ℝ, to guarantee that S(F)\F is non–empty. For F=ℚ the preceding paragraph supplies such an element; for general countable F no proof is known. Conversely nobody has been able to construct a countable field F with S(F)=F, so the problem is open: we do not know whether the chain A₀⊊A₁⊊A₂⊊… is infinite, or whether it stabilises after finitely many steps.

A decisive positive answer (the chain really is infinite) would follow from the still unproved Schanuel conjecture, because Schanuel would imply for instance that for every α∉ℚ the number 2^{1/α} cannot satisfy any equation of the required form whose exponents already lie in the field generated by α. All we can say unconditionally is that the inclusion A₁⊊A₂ is strict; what happens afterwards is open.

4. Are the A_{n} necessarily fields?
   A₁ certainly is, but in general S(F) need not be closed under addition. For example take

 F = ℚ(2^{√2}); then 2^{√2}∈F and the same equation as above shows 2^{1/√2}∈S(F). There is no known way of producing from equations with exponents and coefficients in F a relation that forces 2^{√2}+2^{1/√2} to belong to S(F). In fact it is believed (again on the strength of Schanuel) that S(F) will very rarely be closed under +. Consequently it is not known whether every A_{n} is a field; we only know this for n=0 and n=1.

5. The number e (and π, log 2, …)
   It is also unknown whether e belongs to any A_{n}. All that is known is that e is not algebraic, so e∉A₁. Deciding whether e lies in A₂ already requires settling an outstanding problem in exponential diophantine geometry (more precisely, whether e is “exponentially algebraic’’ over ℚ̄).

Summary.
• A₀⊊A₁=ℚ̄;
• A₁⊊A₂ (witness 2^{1/√2});
• for n≥2 it is an open problem whether A_{n}⊊A_{n+1};
• it is not known whether each A_{n} is a field;
• it is not known whether e is contained in ⋃{n<ω} A{n}.