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Poincare Conjecture ProofH. Vic Dannon5.2∪T = T has open sets G =∪θθ∈[0,2 π]θ∈ΘGθProof:(⊇ ) For each θ ∈ [0,2 π], T ⊇ Tθ.(⊆) If x ∈G∈ T , then x belongs to some θ -angle hyperplane forsome θ [0,2 π]by 4.3. ∩∈ . Hence, ( θ-)x ∈ G the angle hyperplane = G θ∈ T θ25.3 For each θ , f θT θis a topology on S θ, with open sets f θGProof: For each θ ∈ [0,2 π], f θis homeomorphism.θ5.4= ∪3fθ θis a topology on S with open sets ∪ fGθ θθ∈[0,2 π]θ∈Θ∈τT T .Proof:Empty set∅∈Tθ⇒ ∅∈f θ θT ⇒ ∅∈ f∪ θT θ= T .θ∈[0,2 π]2The Space Σθ∈ T ⇒θ∪ ∪ T T .3 2(θ, θ)θ θθ∈[0,2 π] θ∈[0,2 π]S = S ∈ f =Infinite UnionFor any i∈ I , letiG∈T. We’ll show thatiG∈∪ T .i∈IFor each ii∈ I , there is Θ ∈ τ , so that for all θ i ∈Θ, i there areG iθiii∪ . Hence,Gf G∈ T i so that =iθθ iθi iθ ∈Θ14
Poincare Conjecture ProofH. Vic DannoniiG∪ = ∪ ∪ f θiGi.θi ii∈I i∈Iθ ∈ΘTaking the union first over all iSinceiG∪ = ∪ ∪∈ θ∈Θ ∈i I i i Ii∈ I , then over all θ ∈∪ Θ ,∪i∈Iii∪ Θ ∈ τ , and ∪ fθGθ ∈ fθ θi∈Ii∈IFinite intersectionLetSince 1 21 2T . We’ll show that G , G∈ 1 2T , there areG , G∈ fG θiθT , we haveG ∩G ∈ T .i∈IiG∈∪ T .i∈I1 21 1Θ , Θ ∈ τ so that for all θ ∈Θ , and2 2for all θ ∈Θ , there are ∈ T 1 , ∈ T 2 so thatG 1θ1θG 2θ1 12 2= ∪ f θ1G1, and G = f 2 2θθGθ1 12 2Gθ ∈Θ2θ∪ .θ ∈ΘHence,∩ ∪ ∩ ∪GG= f G f G1 2 1 21 1 2θ2θθ θ1 1 2 2θ ∈Θ=θ∈Θθ∈Θ∪ f G ∩1 2∩Θ1 2θ θfGθ θ1 2 1 2Since Θ∩ Θ ∈ τ , and fθGθ∩fGθ θ∈ fθT θ, we have1 2GG∈∩ T .5.5 is Σ3 { θθ ∈ π }T Sup Topology on , and T : [0,2 ] partitions T15
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