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elim的(0)~(5)已被批臭,只是elim脸厚,再批无益。其实由elim给出的递降集列通项的定义 \(N_k:=\{k+1,k+2,…\}\)求\(\displaystyle\bigcap_{k=1}^∞ A_k\)只须两步:
1、验证集合列\(\{A_k\}\)单调递减,并求出通项的极限:
易证:\(A_1\supset A_2\)\(\supset A_3\)\(\supset\)…\(\supset A_k\)\(\supset\)…\(\supset \displaystyle\lim_{k\to\infty}A_k=\displaystyle\lim_{n\to\infty}\{n+1,n+2,n+3,…\}\)
2、根据周民强《实变函数论》定义1.8:设\(\{A_k\}\)是一个集合列,若\(A\supset A_2\)\(\supset A_3\)\(\supset\)…\(\supset A_k\)……,则称此集合列为递减集合列。此时我们称其交集\(\displaystyle\bigcap_{k=1}^∞ A_k\)为集合列\(\{A_k\}\)极限集,记为\(\displaystyle\lim_{n\to\infty} A_k\)写出\(\displaystyle\bigcap_{k=1}^∞ A_k\)\(=\displaystyle\lim_{n\to\infty}\{n+1,n+2,n+3…\}≠\phi\).由于elim不能证明自然数集中哪个自然数n没有后继,故此\(\displaystyle\lim_{n\to\infty}\{n+1,n+2,n+3…\}\)每个数都是客观存在的,所以\(\displaystyle\lim_{n\to\infty}\{n+1,n+2,n+3…\}≠\phi\). |
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