【摘要】 康托集分解为2n个分离闭子集C=C1∪C2∪…C2n,则存在f∶C→C满足f同胚映射,f∶Ci→C2n-1+i x<y∈Ci f(x)<f(y)或x<y x∈Ci y∈Ci f(x)>f(y)i=1,2…2n-1f∶C2n-1+j→Cj x<y x∈C2n-1+j y∈C2n-1+j f(x)<f(y)或f(x)>f(y)j=1,2…2n-1 f∶En→Em n>m≥1 f连续映射.至少有不可数多个反极点Pα-Pαα∈A A不可数.f(Pα)=f(-Pα).更多还原

【Abstract】 Candor set denote C1,C2,…,C2n,Close set disyoiul exist f∶C→C.Hememorphism mapping satisfy f∶Ci x<y∈Ci f(x)<f(y) or x<y x∈Ci y∈Ci f(x)>f(y) i=1,2…2n-1f∶C2n-1+j→Cj x<y x∈C2n-1+j y∈C2n-1+j f(x)<f(y) or f(x)>f(y) j=1,2…2n-1 f∶En→Em n>m≥1 f Contiunus mapping exist not countable antipodal point Pα∈En,-Pα∈En α∈A A no countable f(Pα)=f(-Pα).更多还原