Cantor set$≅\prod_{i=1}^\infty\set{0,1}$ with product topology

hbghlyj

The Cantor set is homeomorphic to infinite product of {0,1}

the product topology of $\{0,1\}^\mathbb{N}$ is generated by the sets of the form $U(N,a)=\{(a_n)_{n=1}^\infty\in\{0,1\}^\mathbb{N}:a_N=a\}$ where $N\in\mathbb{N}$ and $a\in\{0,1\}$, and hence it suffices to show that the preimages of these sets $U(N,a)$ are open in the Cantor set.

How to prove the preimage of $U(N,a)$$$\Set{\sum_{n=1}^\infty \frac{a_n}{3^n}|a_N=2a}$$
is open in the Cantor set

Czhang271828

记 $C$ 为 Cantor 集, 取 $\epsilon <3^{-2N}$, 则
\[
\left(\bigcup_{k\in \mathbb Z}\left(\dfrac{k}{3^{N-1}}+\frac{a_N}{3^N}-\epsilon , \dfrac{k}{3^{N-1}}+\frac{a_N+1}{3^N}+\epsilon\right)\right)\cap C.
\]
是 $U(N,a)$ 的原像, 从而是 $C$ 作为 $\mathbb R$ 通常拓扑的子拓扑空间上的开集.

hbghlyj

$$\Set{\sum_{n=1}^\infty \frac{a_n}{3^n}|a_1=2}=\left[\frac23,\sum_{k=1}^\infty {2\over3^k}\right]\cap C=\left(\sum_{k=1}^\infty {1\over3^k},1\right)\cap C$$

Czhang271828

hbghlyj 发表于 2023-5-29 22:29 $$\Set{\sum_{n=1}^\infty \frac{a_n}{3^n}|a_1=2}=\left[\frac23,\sum_{k=1}^\infty {2\over3^k}\right]\c ...

开区间覆盖例子: $U(4,1)$:

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hbghlyj

Given a collection $ S $ of sets, consider the Cartesian product $ {\textstyle X=\prod _{Y\in S}Y} $ of all sets in the collection. The canonical projection corresponding to some $ Y\in S $ is the function $ p_{Y}:X\to Y $ that maps every element of the product to its $ Y $ component. A cylinder set is a preimage of a canonical projection or finite intersection of such preimages. Explicitly, it is a set of the form, $ {\displaystyle \bigcap _{i=1}^{n}p_{Y_{i}}^{-1}\left(A_{i}\right)=\left\{\left(x\right)\in X\mid x_{Y_{1}}\in A_{1},\dots ,x_{Y_{n}}\in A_{n}\right\}} $ for any choice of $ n $, finite sequence of sets $ Y_{1},...Y_{n}\in S $ and subsets $ A_{i}\subseteq Y_{i} $ for $ 1\leq i\leq n $. Here $ x_{Y}\in Y $ denotes the $ Y $ component of $ x\in X $.

Then, when all sets in $ S $ are topological spaces, the product topology is generated by cylinder sets corresponding to the components' open sets. That is cylinders of the form $ {\textstyle \bigcap _{i=1}^{n}p_{Y_{i}}^{-1}\left(U_{i}\right)} $ where for each $ i $, $ U_{i} $ is open in $ Y_{i} $. In the same manner, in case of measurable spaces, the cylinder σ-algebra is the one which is generated by cylinder sets corresponding to the components' measurable sets.

hbghlyj

What are the cylinder sets in the Cantor space?
These cylinder sets are indeed clopen sets, but finite unions of them are also open (and not always cylinder sets), and of course all clopen subsets are finite unions of them.

Czhang271828

其实 MSE 中的标题有点奇怪. 标题是

Cantor set $\cong\prod_{i=1}^\infty\set{0,1}$ with discrete topology.

我第一时间看到的是

Cantor set $\cong\Big(\prod_{i=1}^\infty\set{0,1}\Big)$ with discrete topology.

看完描述才知道标题的本意是

Cantor set $\cong\prod_{i=1}^\infty\Big(\set{0,1}$ with discrete topology$\Big)$.

其实将标题改写作

Cantor set $\cong\prod_{i=1}^\infty\set{0,1}$ with product topology.

都会比原来的清晰一点.

一般的惯例: 有限点集拓扑默认是离散的, $\prod_{i\in I} X_i$ 默认是积拓扑.

或许论坛里可以优化原标题的表述.

hbghlyj

Czhang271828 发表于 2023-5-31 16:34
一般的惯例: 有限点集拓扑默认是离散的, $\prod_{i\in I} X_i$ 默认是积拓扑.

确实！已修改