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发表于 2025-5-23 22:56
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本帖最后由 春风晚霞 于 2025-5-24 07:24 编辑
由冯\(\cdot\)诺依曼自然数定义(或自然数生成法则)得:\(0=\phi\),\(1=\{0\}\),\(2=\{0,\)\(1\}\),\(3=\{0,1,2\}\),…,\(k=\{0,1,2,…(k-1)\}\),… .
两端分别求并得:\(\displaystyle\bigcup_{k=0}^{\infty}k=\)\(\displaystyle\bigcup_{k=0}^{\infty}\{\phi,0,1,2,3,…,(k-1)\}\) .所以\(\displaystyle\lim_{n \to \infty}\{1,2,…,n\}=\)\(\displaystyle\bigcup_{n=1}^{\infty}\{\phi,0,\)\(1,2,3,…,\)\((n-1)\}\).
若\(\displaystyle\lim_{n \to \infty}n\notin\mathbb{N}\),则\(\mathbb{N}\cap\displaystyle\lim_{n \to \infty}\{\phi,1,2,…n\}=\)
\( \mathbb{N}\cap\displaystyle\bigcup_{n=0}^{\infty}\{\phi,0,1,2,3,…,(n-1)\}=\)\( \displaystyle\bigcup_{n=1}^{\infty}\mathbb{N}\)\(\cap\{\phi,0,1,2,\)\(3,…,(n-1)\}=\phi\).所以\(\mathbb{N}\cap\{\phi\}=\)\(\mathbb{N}\cap\{\phi,0\}=\)\(\mathbb{N}\cap\{\phi,0,1,2\}=\)……\(=\mathbb{N}\cap\{\phi,0,1,2,…(\displaystyle\lim_{n \to \infty}n-1)\}=\phi\),所以\(\mathbb{N}=\phi\),这与\(\mathbb{N}\ne\phi\)矛盾!所以\(\displaystyle\lim_{n \to \infty}n=\infty\in\mathbb{N}\) |
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