文档介绍：习题1111.....(1)()()()CABACBA∩∩∩\\=;(2)()()()CBCACBA\\\∪∪=;(3)()()()CABACBA∩∪\\\=.证明(1))()C\BAA∩∩∩=)()(ccCBAABA∩∩∪∩∩=cCABA)()(∩∩∩=)(\)(CABA∩∩=.(BAA∩∪∪)(C\B)(=)()(ccCBCA∩∪∩==)\()\(CACA∪.(3))(\C)\(BAA∩=ccCBA)(∩∩=)(CBAc∪∩=)()(CABAc∩∪∩=)()\(CABA∩∪=..(1)()ABBA=∪\的充分必要条件是:AB?;(2)()ABBA=\∪的充分必要条件是:=BA∩?;(3)()()BBABBA\\∪∪=的充分必要条件是:=B?.证明(1)====∪∪∩∪∪∩∪)()()()\(的充要条是:.AB?(ccBABBBABBABBA∩∩∪∩∩∪∪===)()()(\)(=\)(∪成立,则ABAc=∩,于是有cBA?,可得.?=BA∩反之若,?≠BA∩取BAx∩∈,则BxAx∈∈且,那么BxAx?∈且与cBA??=BA∩成立,则cBA?,于是有ABAc=∩,即.\)(ABBA=∪(3)\)()\(∪∪=,即.\cCABABA∩∪==若,?≠B取,Bx∈则,cBx?于是,cBAx∩?但,BAx∪∈与cCABA∩∪=?=B成立,显然BABA\=∪成立,即BBABBA\)()\(∪∪=.(1)如果{}nA是渐张集列,即),1(1≥??+nAAnn则{}nA收敛且∪∞=∞→=1;limnnnnAA(2)如果{}nA是渐缩集列,即),1(1≥??+nAAnn则{}nA收敛且∩∞=∞→=(1)设),1(1≥??+nAAnn则对任意∪∞=∈1,nnAx存在N使得,NAx∈从而),(NnAxN≥?∈所以,limnnAx∞→∈∞→∞=?∪又因为∪∞=∞→∞→??1,limlimnnnnnnAAA由此可见{}nA收敛且∪∞=∞→=1;limnnnnAA(2)当)1(1≥??+nAAnn时,对于,limnnAx∞→∈存在)1(1≥?<+knnkk使得),1(≥?∈kAxkn于是对于任意的,1≥n存在0k使得nnk>0,从而,0nnAAxk?∈∩∞=∞→?nnnnAA又因为,limlim1nnnnnnAAA∞→∞→∞=??∩所以可知{}nA收敛且∩∞=∞→=,c为任意实数,证明:(1)??????+≥=>∞=ncfEcfEn1][1∪;(2)??????+<=≤∞=ncfEcfEn1][1∩;(3)若))(()(limExxfxfnn∈?=∞→,则对任意实数c有???????>=???????>=≥∞→∞=∞=∞=∞=kcfEkcfEcfEnnknNnNk1lim1][111∩∩∪∩.证明(1)对任意的[],cfEx>∈有,)(cxf>则存在+∈Zn使得ncxf1)(+≥,1??????+≥∈∪∞=??????+≥∈nncfEx故[];11∪∞=??????+≥?>nncfEcfE另一方面,若,11∪∞=??????+≥∈nncfEx则存在+∈Zn0使得,110∪∞=??????+≥∈cxf>+≥01)(,故[]cfEx>∈.则有[].11∪∞=??????+≥?>nncfEcfE(2)设[]cfEx≤∈,则cxf≤)(,从而对任意的+∈Zn,都有ncxf1)(+<,于是∩∞=??????+<∈11nncfEx,故有[];11∩∞=??????+<?≤nncfEcfE另一方面,设∩∞=??????+<∈11nncfEx,则对于任意的+∈Zn,有ncxf1)(+<,由n的任意性,可知cxf≤)(,即[]cfEx≤∈,故[]∩∞=??????+<?≤11nncfEcfE.(3)设[]cfEx≥∈,则cxf≥)(.由),)(()(limExxfxfnn∈?=∞→可得对于任意的+∈Zk,存在N使得)(1|)()(|Nnkxfxfn≥?<?,即)1(11)()(≥?≥?>kkckxfxfn,即kcxfn1)(?>,故)1(1lim≥????????>∈∞→kkcfExnn,所以∩∞=∞→???????>∈11limknnkcfEx,故[]∩∞=∞→???????>?≥11limknnkcfEcfE;另一方面,设∩∞=∞→???????>∈101limknnkcfEx,则对任意+∈Zk有???????>∈∞→