James Munkres Topology: Sec 37 Exer 1
Exercise 1. Let \(X\) be a space. Let \(\mathcal{D}\) be a collection of subsets of \(X\) that is maximal with respect to the finite intersection property (FIP).
(a) Show that \(x \in \bar{D}\) for every \(D \in \mathcal{D}\) if and only if every neighborhood of \(x\) belongs to \(\mathcal{D}\). Which implication uses maximality of \(\mathcal{D}\)?
Proof:
Prove in the forward direction
Let \(x \in \bar{D}\) for every \(D \in \mathcal{D}\). Let \(U\) be any neighborhood of \(x\) in \(X\). According to Theorem 17.5 (a) in Section 17, if \(x \in \bar{D}\), we have \(U \cap D \neq \varPhi\). This means any neighborhood \(U\) of \(X\) intersects every element in the maximal collection \(\mathcal{D}\). According to Lemma 37.2 (b), \(U \in \mathcal{D}\). The maximality of \(\mathcal{D}\) is used when applying this lemma.
Prove in the backward direction
If there exists a \(D_0 \in \mathcal{D}\) such that \(x \notin \bar{D}_0\), \(x\) belongs to the complement of \(\bar{D}_0\), which is open in \(X\). According to the given condition \(U \in D\) for all \(D \in \mathcal{D}\), \(\bar{D}_0^c\) also belongs to \(\mathcal{D}\). Then \(\bar{D}_0^c \cap D_0 = \varPhi\) contradicts the fact that \(\mathcal{D}\) has the FIP.
(b) Let \(D \in \mathcal{D}\). Show that if \(A \supset D\), then \(A \in \mathcal{D}\).
Proof: Because \(\mathcal{D}\) has the FIP, for all \(D' \in \mathcal{D}\), \(D \cap D' \neq \varPhi\). Because \(D\) is contained in \(A\), \(A \cap D' \neq \varPhi\). According to Lemma 37.2 (b), \(A \in \mathcal{D}\).
(c) Show that if \(X\) satisfies the \(T_1\) axiom, there is at most one point belonging to the intersection of all elements in \(\mathcal{D}\), i.e., \(\bigcap_{D \in \mathcal{D}} \bar{D}\).
Proof: Assume that there are at least two points \(x_1\) and \(x_2\) in \(\bigcap_{D \in \mathcal{D}} \bar{D}\). If \(X\) is a Hausdorff space, there are disjoint open sets \(U_1\) and \(U_2\) in \(X\) containing \(x_1\) and \(x_2\) respectively. According to part (a) of this exercise, we have \(U_1 \in \mathcal{D}\) and \(U_2 \in \mathcal{D}\). Then, \(U_1\) and \(U_2\) being disjoint contradicts the fact that \(\mathcal{D}\) has the FIP.
Unfortunately, the given condition in this exercise, i.e. \(X\) satisfies the \(T_1\) axiom, is weaker than the above assumption that \(X\) is Hausdorff. Hence the above proof does not work. However, there seems no obvious or direct proof for the claim in the exercise. This may imply that the original statement is erroneous.
According to the discussion here, a counter example involving the cofinite topology \(\mathcal{T}_c\) on the set of natural numbers \(\mathbb{N}\) is given. It further shows that the intersection of all the elements in the maximal collection \(\mathcal{D}\) is actually \(\mathbb{N}\) itself. This contradicts the claim in the exercise. In the following, the construction of the counter example will be given.
Definition of the cofinite topology
Definition (Cofinite topology) Let \(\mathcal{T}_c\) be the cofinite topology of the space \(X\). Then for all \(U \in \mathcal{T}_c\), either \(U\) is empty or its complement \(U^c\) is finite.
Next, we'll show \(\mathcal{T}_c\) satisfying the conditions in the above definition really defines a topology on \(X\).
It is obvious that \(\varPhi\) belongs to \(\mathcal{T}_c\).
When \(U = X\), \(U^c = \varPhi\), which is finite. Hence \(X\) belongs to \(\mathcal{T}_c\).
Check the closeness of the union operation.
Let \(\{U_i\}_{i \in I}\) be a collection of open sets in \(\mathcal{T}_c\). If some \(U_i\) in the collection is empty, it has no contribution to the union. Hence we assume all the \(U_i\) in the collection are non-empty.
Then we have
\[
\left( \bigcup_{i \in I} U_i \right)^c = \bigcap_{i \in I} U_i^c,
\]
where each \(U_i^c\) is finite. The above intersection of \(\{U_i^c\}_{i \in I}\) is a subset of finite set, which is also finite. Therefore \(\bigcup_{i \in I} U_i \in \mathcal{T}_c\).Check the closeness of the finite intersection operation.
For a finite collection of open sets in \(\mathcal{T}_c\), we have
\[
\left( \bigcap_{k = 1}^n U_k \right)^c = \bigcup_{k = 1}^n U_k^c.
\]
Because each \(U_i^c\) is a finite set, the union of a finite number of finite sets is still finite. Hence \(\bigcap_{k = 1}^n U_k \in \mathcal{T}_c\).
Due to the above analysis, \(\mathcal{T}_c\) is really a topology for \(X\). We also know that because every finite set in \(X\) assigned with the topology \(\mathcal{T}_c\) is closed, \(X\) satisfies the \(T_1\) axiom.
Counter example derived from the cofinite topology on \(\mathbb{N}\)
Let the set of natural numbers \(\mathbb{N}\) be assigned with the cofinite topology \(\mathcal{T}_c\). \(\mathbb{N}\) satisfies the \(T_1\) axiom. Let \(\mathcal{C}\) be a collection of all those subsets in \(\mathbb{N}\), each of which has a finite complement. This means all the open sets in \(\mathcal{T}_c\) except \(\varPhi\) are included in \(\mathcal{C}\). Accordingly, the following can be obtained.
For all \(U \in \mathcal{C}\), because \(\mathbb{N} - U\) is finite while \(\mathbb{N}\) is infinite, \(U\) is an infinite subset of \(\mathbb{N}\).
Let \(\{U_k\}_{k = 1}^n\) be a finite collection arbitrarily selected from \(\mathcal{C}\). Then we have
\[
\left( \bigcap_{k = 1}^n U_k \right)^c = \bigcup_{k = 1}^n U_k^c.
\]
Because \(U_k^c\) for each \(k\) from \(1\) to \(n\) is a non-empty finite set, their finite union is still finite. Because \(\mathbb{N}\) is infinite, \(\bigcap_{k = 1}^n U_k\) must be infinite, which is a non-empty open set. Therefore \(\mathcal{C}\) has the FIP.
Next, by applying the Zorn's Lemma, a maximal collection \(\mathcal{D}\) exists, which contains \(\mathcal{C}\) as its sub-collection and also has the FIP. For all \(D \in \mathcal{D}\), \(D\) must have infinite number of elements. Otherwise, if \(D = \{d_i\}_{i = 1}^m\), we can select a sub-collection \(\{C_i\}_{i = 1}^m\) from \(\mathcal{C}\), such that \(d_i \notin C_i\). Then \(D \cap C_1 \cap \cdots \cap C_m = \varPhi\), which contradicts the fact that \(\mathcal{D}\) has the FIP.
Select an arbitrary \(x\) in \(D^c\), for any open set \(U\) in \(\mathcal{T}_c\) containing \(x\), it has non-empty intersection with \(D\) because \(D\) is an infinite set. This means any point \(x\) in \(D^c\) is a limiting point of \(D\), so \(\bar{D} = D \cup D^c = \mathbb{N}\). Hence \(\bigcap_{D \in \mathcal{D}} \bar{D} = \mathbb{N}\), which obviously has more than one point.
James Munkres Topology: Sec 37 Exer 1的更多相关文章
- James Munkres Topology: Sec 18 Exer 12
Theorem 18.4 in James Munkres “Topology” states that if a function \(f : A \rightarrow X \times Y\) ...
- James Munkres Topology: Sec 22 Exer 6
Exercise 22.6 Recall that \(\mathbb{R}_{K}\) denotes the real line in the \(K\)-topology. Let \(Y\) ...
- James Munkres Topology: Sec 22 Exer 3
Exercise 22.3 Let \(\pi_1: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}\) be projection on th ...
- James Munkres Topology: Sec 22 Example 1
Example 1 Let \(X\) be the subspace \([0,1]\cup[2,3]\) of \(\mathbb{R}\), and let \(Y\) be the subsp ...
- James Munkres Topology: Lemma 21.2 The sequence lemma
Lemma 21.2 (The sequence lemma) Let \(X\) be a topological space; let \(A \subset X\). If there is a ...
- James Munkres Topology: Theorem 20.3 and metric equivalence
Proof of Theorem 20.3 Theorem 20.3 The topologies on \(\mathbb{R}^n\) induced by the euclidean metri ...
- James Munkres Topology: Theorem 20.4
Theorem 20.4 The uniform topology on \(\mathbb{R}^J\) is finer than the product topology and coarser ...
- James Munkres Topology: Theorem 19.6
Theorem 19.6 Let \(f: A \rightarrow \prod_{\alpha \in J} X_{\alpha}\) be given by the equation \[ f( ...
- James Munkres Topology: Theorem 16.3
Theorem 16.3 If \(A\) is a subspace of \(X\) and \(B\) is a subspace of \(Y\), then the product topo ...
随机推荐
- vc++基础班[27]---实现一个简单的任务管理器
因为任务管理器中涉及到进程的枚举操作,所以把两节课的知识点合并到一起来讲! ①.设计界面.以及列表控件变量的绑定: ②.列表控件样式的指定: m_TaskList.SetExtendedSty ...
- sqlite处理数据
# coding: UTF-8 import platform from _utils.patrol2 import run_cmd, data_format, report_format impor ...
- 题解-PKUWC2018 随机算法
Problem loj2540 题意简述:给定\(n\)个点的无向图,给定求最大独立集的近似算法:随机排列\(1\cdots n\),按照该排列顺序贪心构造最大独立集(即对每个点能加入独立集就加),求 ...
- C++面向对象的特点
C++面向对象的特点 面向对象的特点主要有: 封装, 继承, 多态; 现在自己的简单理解如下, 但要明白具体怎么实现, 背后的原理是什么? 什么是封装, C++怎么实现封装 封装的大致可以分为: 函数 ...
- 全系列Unity4.x.x到2017.1.1破解Win&Mac!最新Unity2017.1.1p3&4.7.2f1破解!
Unity官网所有版本下载地址请戳: http://unity3d.com/unity/download/archive 补丁版本请戳: http://unity3d.com/cn/unity/qa/ ...
- button 去掉原生边框
button按钮触发 hover 时,自带边框会显示,尤其是 button 设置圆角时,如图: 解决办法: outline: 0;
- $Django 路飞学城项目简介
- 基于极验实现动态验证码 - 在线视频播放:cc,HTML用的Flash - 基于Rest Framework实现 API接口 - 自定义rest认证token 认证 - 序列化以及自定义验证对请求 ...
- python操作三大主流数据库(7)python操作mongodb数据库①mongodb的安装和简单使用
python操作mongodb数据库①mongodb的安装和简单使用 参考文档:中文版:http://www.mongoing.com/docs/crud.html英文版:https://docs.m ...
- mysql 命令行常用命令
1.显示数据库列表. show databases; 2.显示库中的数据表: use mysql; show tables; 3.显示数据表的结构: describe 表名; 4.建库: cr ...
- git命令(版本控制之道读书笔记)
1.在Windows中安装完git后,需要进行一下配置:$ git config --global user.name "zhangliang"$ git config --glo ...