Β· 2,467 chars Β· 8 deps Β· depth 22 Reason: First publication of the proof: beyond the maximum of the four indices supplied by the four convergences, all four arguments lie within the modulus supplied by continuity.
Take the single modulus supplied by continuity at the quadruple and the four indices supplied by the four convergences; beyond their maximum all four arguments are within the modulus.
also satisfy βF(y,s,q,Y)βF(x0β,r0β,p0β,X0β)β<Ξ΅.
By convergence of the four given sequences there are N1β,N2β,N3β,N4ββN such that: dEβ(ykβ,x0β)<Ξ΄ for every kβ₯N1β; dRβ(skβ,r0β)=β£skββr0ββ£<Ξ΄ for every kβ₯N2β; dEβ(qkβ,p0β)=β₯qkββp0ββ₯<Ξ΄ for every kβ₯N3β; and dS(n)β(Ykβ,X0β)<Ξ΄ for every kβ₯N4β.
Put N=max{max{N1β,N2β},max{N3β,N4β}}. By claim 1 of Elementary Properties of the Maximum of Two Elements, N1ββ€max{N1β,N2β}β€N and likewise for N2β, N3β and N4β, so Niββ€N for i=1,2,3,4 by the transitivity of claim 1 of Properties of the Order on the Natural Numbers. Let kβN satisfy kβ₯N. Then kβ₯Niβ for each i, again by transitivity, so all four displayed bounds hold with y=ykβ, s=skβ, q=qkβ and Y=Ykβ, and therefore