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发表于 2024-8-13 11:18
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本贴旨在证明 \([0,1]\)与\(\mathscr{P}(\mathbb{N})\) 对等。用康托幂集定理证明\([0,1]\)不可数.
令 \(\mathscr{L}(\mathbb{N}_+)=\{A\in\{B,B^c\}:\;B\subset\mathbb{N},\;0< |B|\in\mathbb{N}_+\}\) 易见 \(\mathscr{L}(\mathbb{N_+})\) 可数。
\(\quad\)\(\bigg(A\mapsto \displaystyle\sum_{n\in\mathbb{N}_+}2^n\chi_A(n) \) 是\(\mathbb{N}_+\)的有限子集到\(\mathbb{N}\) 的单射.\(\bigg)\)
令 \(C_0 = \displaystyle\{{\small\sum_{k=1}^\infty\frac{\chi_A(k)}{2^k}}\mid A\in\mathscr{L}(\mathbb{N}_+)\},\;C=[0,1)-C_0\)
\(\quad\)对 \(\alpha\in C,\;\;a_k=\lfloor 2^k\alpha\rfloor -2\lfloor 2^{k-1}\alpha\rfloor,\;(k=1,2,3,\ldots)\),
\(\quad\)由 \(2^{n-1}\alpha=\lfloor 2^{n-1}\alpha\rfloor+\beta\) 得 \(\lfloor 2^k\alpha\rfloor -2\lfloor 2^{k-1}\alpha\rfloor=\lfloor 2\beta\rfloor\in\{0,1\}\)
\(\quad\)且 \(\displaystyle\sum_{n=1}^\infty\frac{a_n}{2^{n}}=\lim_{m\to\infty}\sum_{n=1}^m\big(\frac{\lfloor 2^n\alpha\rfloor}{2^n}-\frac{\lfloor 2^{n-1}\alpha\rfloor}{2^{n-1}}\big)\)
\(\qquad\displaystyle =\lim_{n\to\infty}\frac{\lfloor 2^n\alpha\rfloor}{2^n} =\lim_{n\to\infty}\frac{2^n\alpha-(2^n\alpha-\lfloor 2^n\alpha\rfloor) }{2^n} = \alpha\)
\(\therefore\quad \alpha\in C\) 与 \(A=\{n\in\mathbb{N}_+:\;\lfloor 2^n\alpha\rfloor -2\lfloor 2^{n-1}\alpha\rfloor = 1\}\in\mathscr{P}(\mathbb{N}_+)-\mathscr{L}(\mathbb{N}_+)\)
\(\qquad\)的关系是1-1对应. 故\(|\mathbb{R}|=|C|=|\mathscr{P}(\mathbb{N})-\mathscr{L}(\mathbb{N}_+)|=|\mathscr{P}(\mathbb{N})|=2^{\aleph_0}>\aleph_0\)
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发表于 2024-8-11 15:05
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