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证明:反证法,假设$X$是完备的度量空间,且是第一纲的,下面我们推出矛盾

   由于$X$是第一纲的,则$X$可表示为可数个疏朗集的并,不妨设$X = igcuplimits_{n = 1}^infty  {{M_n}} $,其中${M_n}$均为疏朗集

任取一个闭球$B(a,1)$,由${M_1}$是疏朗集知,存在$X$中的非空闭球$Bleft( {{a_1},{r_1}} ight) subset Bleft( {a,1} ight)$,使得[Bleft( {{a_1},{r_1}} ight) cap {M_1} = emptyset ]其中不妨设$0<{r_1}<1$,又由${M_2}$是疏朗集知,存在$X$中的非空闭球$Bleft( {{a_2},{r_2}} ight) subset Bleft( {{a_1},{r_1}} ight) $,使得[Bleft( {{a_2},{r_2}} ight) cap {M_2} = emptyset ]其中不妨设$0 < {r_2} < frac{1}{2}$,如此可以选得一套非空闭球[Bleft( {{a_1},{r_1}} ight) supset Bleft( {{a_2},{r_2}} ight) supset  cdots  supset Bleft( {{a_n},{r_n}} ight) supset  cdots ]使得$Bleft( {{a_n},{r_n}} ight) cap {M_n} = emptyset $,且$0 < {r_n} < frac{1}{n}$

   于是由闭球套定理知,存在唯一的$x in igcaplimits_{n = 1}^infty  {Bleft( {{a_n},{r_n}} ight)} $,又$Bleft( {{a_n},{r_n}} ight) cap {M_n} = emptyset $,所以有$x otin igcuplimits_{n = 1}^infty  {{M_n}} $,而$X = igcuplimits_{n = 1}^infty  {{M_n}} $,出现矛盾

原文地址:https://www.cnblogs.com/ly758241/p/3813224.html