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:

  1. 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.

  2. 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 \(X\).

  1. It is obvious that \(\varPhi\) belongs to \(\mathcal{T}_c\).

  2. When \(U = X\), \(U^c = \varPhi\), which is finite. Hence \(X\) belongs to \(\mathcal{T}_c\).

  3. 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 \(\bigcup_{i \in I} U_i \in \mathcal{T}_c\).

  4. 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.

  1. For all \(U \in \mathcal{C}\), because \(\mathbb{N} - U\) is finite while \(\mathbb{N}\) is infinite, \(U\) is an infinite subset of \(\mathbb{N}\).

  2. 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

  1. James Munkres Topology: Sec 18 Exer 12

    Theorem 18.4 in James Munkres “Topology” states that if a function \(f : A \rightarrow X \times Y\) ...

  2. James Munkres Topology: Sec 22 Exer 6

    Exercise 22.6 Recall that \(\mathbb{R}_{K}\) denotes the real line in the \(K\)-topology. Let \(Y\) ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. James Munkres Topology: Theorem 20.4

    Theorem 20.4 The uniform topology on \(\mathbb{R}^J\) is finer than the product topology and coarser ...

  8. 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( ...

  9. 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

  1. Java文件复制

    主要是工作代码,无解释. /** * 将文件或文件夹source复制到dest * <br>目标文件检测: * <br> a.当文件不存在时:需要创建文件 * <br&g ...

  2. centos6下安装php7的memcached扩展

    安装php7的memcached扩展 .编译安装libmemcached- wget https://launchpadlibrarian.net/165454254/libmemcached-1.0 ...

  3. 解决mysql 主从数据库同步不一致的方法

    接着上文 配置完Mysql 主从之后,在使用中可能会出现主从同步失败的情况. mysql> show slave status\G Slave_IO_Running: Yes Slave_SQL ...

  4. Mysql 通过frm&ibd 恢复数据

    mysql存储在磁盘中,各种天灾人祸都会导致数据丢失.大公司的时候我们常常需要做好数据冷热备,对于小公司来说要做好所有数据备份需要支出大量的成本,很多公司也是不现实的.万一还没有做好备份,数据被误删除 ...

  5. SpringBoot中常见注解含义总结

    @RestController @RestController被称为一个构造型(stereotype)注解.它为阅读代码的开发人员提供建议.对于Spring,该类扮演了一个特殊角色.它继承自@Cont ...

  6. Python-socketserver实现并发- 源码分析

    基于tcp的套接字,关键就是两个循环, 一个链接循环,一个通信循环 socketserver模块中分两大类: server类(解决链接问题)和request类(解决通信问题) server类: req ...

  7. Android广播机制

    原文出处: Android总结篇系列:Android广播机制 1.Android广播机制概述 Android广播分为两个方面:广播发送者和广播接收者,通常情况下,BroadcastReceiver指的 ...

  8. LuoGu P2420 让我们异或吧

    其实......这就是个SB题,本来看到这个题,和树上路径有关 于是--我就欣喜地打了一个树剖上去,结果嘞,异或两遍等于没异或 所以这题和LCA屁关系都没有,所以这题就是个树上DFS!!!! 所以它为 ...

  9. fatal: refusing to merge unrelated histories

    Git 提交代码时遇到冲突了,所以 git pull 拉不下来远程代码.使用一下命令解决: git pull origin master --allow-unrelated-histories 然后解 ...

  10. Oracle SQL高级编程——分析函数(窗口函数)全面讲解

    Oracle SQL高级编程--分析函数(窗口函数)全面讲解 注:本文来源于:<Oracle SQL高级编程--分析函数(窗口函数)全面讲解> 概述 分析函数是以一定的方法在一个与当前行相 ...