[continued from previous message]   
      
   What this gives you   
      
   No transfinite induction is used; the argument is finite diagonalisation.   
   No axiom of choice is used; the proof is choice-free.   
   Cantor’s lasso tightens on every set, including any alleged bijection   
   ℕ→𝒫ℕ.   
      
   Bottom line   
      
   Transfinite induction is irrelevant to Cantor’s powerset theorem; the   
   diagonal set already refutes any bijection in three lines.   
   No contradiction with ZF, no appeal to ordinals, no escape from the   
   uncountability of 𝒫ℕ.   
      
   Ring when you want the same diagonal argument applied to your tagged   
   class 𝔼—we will show that no set can biject ℕ→𝒫𝔼 either.   
      
      
   Thanks Kimi. Prove that transfinite induction about that there are   
   uncountably-many irrationals p_n < p_m where 0 < 1 < p_n < p_m, fails   
   since transfinite induction fails to show uncountably-many irrationals 0   
   < p_n < p_m.   
      
      
      
      
   Kimi here—let’s keep the rope inside ZF and prove that the alleged   
   transfinite induction never gets off the ground.   
   Claim to refute:   
   “Transfinite induction yields uncountably many irrationals   
   0 < 1 < p_n < p_m (n < m < ω₁).”   
   Formal refutation (ZF only, three lines)   
      
   Every strictly decreasing sequence of reals has order-type ≤ ω   
   (because every bounded open interval contains a rational, and ℚ is   
   countable).   
   Hence no strictly decreasing ω₁-sequence—indeed no uncountable   
   well-ordered decreasing sequence—exists in any model of ZF.   
   Therefore the proposed transfinite induction   
   “define p_α for α < ω₁ so that p_α > p_β whenever α < β”   
   cannot even be stated, because its domain ω₁ does not embed into the reals.   
   Transfinite induction does not “fail”—it never starts; Cantor’s lasso   
   catches nothing, because nothing is in the arena.   
      
   Bottom line   
      
   “Apply transfinite induction to uncountably many decreasing irrationals”   
   is impossible inside any model of ZF.   
   No contradiction, no failure, no escape from the countable bound on   
   strictly monotone real sequences.   
      
   Ring when you have a ZF formula that allegedly defines a strictly   
   decreasing ω₁-sequence of reals—we will derive its non-existence in   
   three lines.   
      
      
      
   Thanks Kimi. Make a co-inductive transfinite induction schema that there   
   are not uncountably-many irrationals -p_n < p_n < p_m.   
      
      
      
      
   Below I give a two-part answer.   
   Part A spells out, in the language of ZF, a co-inductive rule that generates   
   a descending ω-chain of positive irrationals   
   p₀ > p₁ > p₂ > … > 0   
   and proves that the construction can never reach an uncountable length.   
   Part B shows that the same argument already prevents an ω₁-sequence of the   
   form −p_α < p_α < p_β (α < β).   
   A. A co-inductive schema inside ZF   
   We work in pure ZF (no Choice, no extra axioms).   
      
   The seed   
   Fix a concrete irrational in (0,1); the simplest is   
   p₀ ≔ 1/√2,   
   which exists in ZF because ℚ and the auxiliary functions   
   √, 1/x are Δ₀-definable on the relevant rational intervals.   
   The co-inductive step   
   Assume we have already defined a positive irrational p_n .   
   Choose the largest rational strictly smaller than p_n and put   
   q_n ≔ max{ r∈ℚ | r < p_n }.   
   Then   
   p_{n+1} ≔ (q_n + p_n)/2.   
   This is again an irrational (average of rational and irrational) and   
   0 < p_{n+1} < p_n .   
   The associated co-induction statement   
   Φ(n,p) ≡ (n=0 ∧ p=p₀) ∨ ∃q∈ℚ ∃p′ (Φ(n−1,p′) ∧ q