cantor集为什么是不可数集合

2025-03-09 22:05:35
推荐回答(1个)
回答(1):

不是可数集。将0到1之间的实数用三进制表示,可以知道去掉的是数位含有1的三进制数,剩下的位数只有0和2的三进制数就是康托集,和0到1中的实数的二进制数存在一一对应。又因为0到1的实数不可数,所以康托集不可数~

豆瓣上的相关讨论:

Cantor set为什么是不可数的?

来自: [已注销] 2011-09-30 22:15:54
从Cantor set的构造来看,由闭区间套定理,他就是孤立点集啊。。。而且是有理数的一个子集。。。所以应该是可数的吧,为什么会是不可数呢?

我的想法哪里出了问题呢,求指教,谢谢。
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阿狄 (晴川历历,芳草萋萋) 2011-09-30 22:21:11
怎么有理数了?
其实你可以把Cantor Set的数表示成3进制,则小数点后只有“0”“2”没有“1”。“0”“2”和二进制的“0”“1”是可以一一对应的。
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xdotzzzzzzzzzz (El Psy Congroo) 2011-09-30 22:21:21
cantor set is perfect,and nonempty perfect set is uncountable...
//刚在rudin的书上看到...
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余妙哉 2011-09-30 22:21:40
从三元数列的角度考虑
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[已注销] 2011-09-30 22:24:39
三进制那个我也知道,可就是我这样想哪里错了?它的分点都是有理数啊。
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[已注销] 2011-09-30 22:25:30
我也对这个很纠结
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[已注销] 2011-09-30 22:26:18
按照构造的话,感觉就是区间端点,而区间是可数的,所以端点也应该可数呀
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lethe 2011-09-30 22:31:01
康托集是有理数的子集??你怎么看出来的?分点是有理数,但是分点附近还有没被挖走的点啊
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阿狄 (晴川历历,芳草萋萋) 2011-09-30 22:35:31
康托尔集并不是只有分点啊
比如0.20220222022220...在康托尔集中但不是有理数
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[已注销] 2011-09-30 22:36:19
我知道我又意淫了,仔细思考一下
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[已注销] 2011-09-30 22:40:08
可是cantor集构造的时候留下的都是端点呀,不是吗
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余妙哉 2011-09-30 22:45:39
留下的都是端点?你验证1/4是否在康托集中?如果不在,你能说清是哪一步把这个点删去了吗?
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always waiting (always waiting) 2011-09-30 23:05:29
额,实变函数完全忘了啊。。。。
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[已注销] 2011-10-01 09:33:05
很形象的解释是:每次操作后选取的那些小区间的中点显然在集合中。因此第n次操作导致有[;2^n;]个点被放入集合中,操作是可数次的,也就是aleph 0,因此这些操作导致的点具有基数aleph 1,因此根据这种基数的推导我们知道Cantor set中的元素的基数是aleph 1,因此不可数。
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J.-J. Jiang 2011-10-02 19:12:01
沿LZ的思路:根据区间套定理,每个区间套唯一确定了一个点,而区间套的长度全为 aleph_0,故不同的区间套数目为 2^aleph_0,因此 Cantor set 与连续统等势。
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钱塘新泥 (当回忆重来,相信天仍会很晴.) 2011-10-02 19:38:20
Cantor集不是孤立点集,恰恰相反,它的每个点都是它的聚点.
而且,它的聚点也都在本身中,
也就是说,Cantor集=Cantor集的导集等于自身的导集的集合称为完全集.Cantor集就是一个完全集的例子.
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[已注销] 2011-10-09 19:31:46
闭区间套定理用的怎么不对了?
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[已注销] 2011-10-09 19:32:50
2011-10-02 19:12:01 Triple.J

沿LZ的思路:根据区间套定理,每个区间套唯一确定了一个点,而区间套的长度全为 aleph_0,故不同的区间套数目为 2^aleph_0,因此 Cantor set 与连续统等势。

==

这个?
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[已注销] 2011-10-10 00:05:20
对,就是Triple.J 那个。
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[已注销] 2011-10-10 21:49:01
嗯嗯嗯,明白了,谢谢大家

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