User blog:Rgetar/Proof that limit of epsilon numbers is epsilon number

I proved it in 2018, then forgot, then proved again, then forgot.

Limits of sequences of ordinals with some property
Generally, if ordinals of some sequence have some property, then limit of this sequence does not always has this property.

Examples: So why we do know that limit of ε0, ε1, ε2, ε3, ε4, ... is ε too (εω)? Maybe, this limit is some non-epsilon ordinal α, and εω is limit of α, ωα, ωω α, ωω ω α , ωω ω ω α   , ...?
 * elements of sequence 0, 1, 2, 3, 4, ... are finite, but its limit ω is infinite
 * elements of sequence 1, 2, 3, 4, 5, ... are successors, but its limit ω is limit
 * elements of sequence ω, Ω, Ω2, Ω3, Ω4, ... are regular, but its limit Ωω is singular

Proof
By definition, epsilon number is ordinal εx such as Find ωα. According ordinal arithmetic, α is a limit ordinal, and α is limit of so ωα is limit of that is ωα is limit of that is So α is epsilon number The subscript is ω, since if it was less than ω, that is finite, then α would be less than some elements of ε0, ε1, ε2, ε3, ε4, ..., and if it was larger than ω, then there would be lesser ordinal εω larger than any element of this sequence (the same proof is for any other sequence of subscripts).
 * ωεx = εx
 * if γ is a limit ordinal, then βγ is the limit of the βδ for all δ < γ
 * ε0, ε1, ε2, ε3, ε4, ...,
 * ωε0, ωε1, ωε3, ωε3, ωε4, ...,
 * ε0, ε1, ε2, ε3, ε4, ...,
 * ωα = α
 * α = εω

So any limit of epsilons is also epsilon: where
 * sup{εα[0], εα[1], εα[2], εα[3], εα[4], ...} = εα,
 * α = sup{α[0], α[1], α[2], α[3], α[4], ...}

Zetas
What about zetas?

By definition ζx is ordinal such as that is any zeta is also epsilon. So limit of zetas is epsilon, which subscript is limit of these zetas. Let then Similarly, where
 * εζ x = ζx
 * let α = sup{ζα[0], ζα[1], ζα[2], ζα[3], ζα[4], ...}
 * sup{ζα[0], ζα[1], ζα[2], ζα[3], ζα[4], ...} = sup{εζ α[0], εζ α[1] , εζ α[2] , εζ α[3] , εζ α[4] , ...} = εα = α
 * α = εα = ζx
 * x = α
 * sup{ζα[0], ζα[1], ζα[2], ζα[3], ζα[4], ...} = ζα,
 * α = sup{α[0], α[1], α[2], α[3], α[4], ...}

Beyond
I remember that I proved the same for other functions including φ(ω, x) and perhaps Γx (I'm not sure about the latter).