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发表于 2025-8-19 02:45
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\(\huge\color{green}{\textbf{孬种反数学猿声啼不住, 滚驴离正道已隔万重山}}\)
【定理】\(\lim n=\sup \mathbb{N}\)
【证明】对\(m< n\)令\(n\small\to\infty\)得\(m< \lim n\small(\forall m\in\mathbb{N})\)
\(\qquad\)即\(v=\lim n\)是\(\mathbb{N}\)的上界. 设\(\mu\)是\(\mathbb{N}\)的上界, 则
\(\qquad\)又有\(n<\mu\small(\forall n\in\mathbb{N})\). 再令\(\small n\to\infty\)得\(\lim n\le \mu\)
\(\qquad\)可见\(v\)是\(\mathbb{N}\)的最小上界. 即 \(\boxed{\lim n=\sup \mathbb{N}}\)
【推论】\(\lim n\not\in \mathbb{N}\)
【证明】因为\(\mathbb{N}\)无大元, 故 \(\lim n=\sup\mathbb{N}\not\in \mathbb{N}\).
\(\;\,\)本贴起底孬种回避 \(\lim n=\sup \mathbb{N}\) 的原因:
\(\;\,\)这直接导致 \(\lim n\not\in \mathbb{N}\)!
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