# 数学代写|数学分析代写Mathematical Analysis代考|Metrization

#### Doug I. Jones

Lorem ipsum dolor sit amet, cons the all tetur adiscing elit

couryes™为您提供可以保分的包课服务

couryes-lab™ 为您的留学生涯保驾护航 在代写数学分析Mathematical Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数学分析Mathematical Analysis代写方面经验极为丰富，各种代写数学分析Mathematical Analysis相关的作业也就用不着说。

## 数学代写|数学分析代写Mathematical Analysis代考|Metrization

We now turn to the question of which topologies are induced by a metric. Theorem 5.11.3 is the main result in this section. Although it is not the best known result, it does establish sufficient conditions for metrization. The proof techniques we develop along the path to theorem 5.11 .3 are elegant and important in their own right. We first state the following definition.

Definition. A topological space $(X, \mathcal{J})$ is metrizable if there is a metric $d$ on $X$ that induces the topology $\mathcal{T}$.

Lemma 5.11.1. Suppose $X$ is a normal space, and let $E$ and $F$ be disjoint closed subsets of $X$. Let $C$ be the set of rational points in the interval $[0,1]$. Then there exists a countable collection of open subsets $\left{U_p: p \in C\right}$ such that
if $p, q \in C$ and $p<q$, then $\bar{U}p \subseteq U_q$. Additionally, for all $p \in C, E \subseteq U_p$, and $\bar{U}_p \subseteq X-F$ Proof. Let $p_0=0$, and $p_1=1$, and let $\left{p_2, p_3, p_4, \ldots\right}$ be an enumeration of the rational point in $(0,1)$. Since $E \subseteq X-F$, theorem 5.6 .3 yields an open set $U_1$ such that $E \subseteq U_1 \subseteq \bar{U}_1 \subseteq X-F$. Another application of theorem 5.6.3 yields an open set $U_0$ such that $E \subseteq U_0 \subseteq \bar{U}_0 \subseteq U_1$. The rest of the construction is inductive. Suppose that, for each element $p_i$ of the finite set $C_n=\left{p_0, \ldots, p_n\right}$, we have found an open set $U{p_i}$ such that the sets $U_{p_1}, \ldots, U_{p_n}$ satisfy condition $\left(^*\right)$ for $p, q \in C_n$. Consider the rational number $p_{n+1}$. It must fall strictly between two elements of $C_n$, say, $p_i<p_{n+1}<p_j$. Again by theorem 5.6.3, there exists an open set $U_{p_{n+1}}$ such that $\bar{U}{p_i} \subseteq U{p_{n+1}} \subseteq \bar{U}{p{n+1}} \subseteq U_{p_j}$. By construction, the sets $U_{p_0}, \ldots, U_{p_{n+1}}$ satisfy condition ( $\left.{ }^*\right)$ for $p, q \in C_{n+1}$. Since, for every pair of points $p$ and $q$ in $C$, there is a finite set $C_n$ that contains $p$ and $q$, the proof is complete.

The inclusions $E \subseteq U_p$, and $\bar{U}_p \subseteq X-F$ for all $p \in C$ are obvious since $E \subseteq U_0$ and $\bar{U}_1 \subseteq X-F$

## 数学代写|数学分析代写Mathematical Analysis代考|The Product of Infinitely Many Spaces

This section generalizes section 5.4. First we review some terminology and notation.

Let $\left{X_\alpha\right}_{\alpha \in I}$ be an arbitrary collection of nonempty sets. The Cartesian product $X=\prod_{\alpha \in I} X_\alpha$ is the set of all functions $x: I \rightarrow \cup_{\alpha \in I} X_\alpha$ such that, for every $\alpha \in I$, $x(\alpha) \in X_\alpha$. We write $x_\alpha$ instead of $x(\alpha)$, and we denote an element of $X$ by $x=\left(x_\alpha\right){\alpha \in I}$, or simply $x=\left(x\alpha\right)$. For a fixed $\alpha \in I$, the projection of $X$ onto the factor set $X_\alpha$ is the function $\pi_\alpha(x)=x_\alpha$.

Let $\left{\left(X_\alpha, \mathcal{J}\alpha\right)\right}{\alpha \in I}$ be a collection of topological spaces, and let $X=\prod_\alpha X_\alpha$ be the Cartesian product of the underlying sets. As in the definition of the product topology in section 5.4 , we would like the product topology to guarantee the continuity of all the projections $\pi_\alpha: X \rightarrow X_\alpha$. One might be tempted to adopt the following simple generalization of the product of finitely many spaces. Consider the topology $\mathcal{T}$, which has the following subbase:
$$\left{\prod_{\alpha \in I} U_\alpha: \alpha \in I, U_\alpha \in \mathcal{J}\alpha\right} .$$ It would be a hasty decision to define $\mathcal{T}$ to be the product topology. Although $\mathcal{T}$ certainly guarantees the continuity of all the projections, it is too wasteful because, in order to guarantee the continuity of $\pi\alpha$, we only need the openness of sets of the form $\pi_\alpha^{-1}\left(U_\alpha\right)$, where $U_\alpha \in \mathcal{T}\alpha$. A little reflection shows that $$\pi\alpha^{-1}\left(U_\alpha\right)=U_\alpha \times \prod_{\beta \neq \alpha} X_{\beta \cdot}{ }^6$$
Therefore the smallest topology which guarantees the continuity of all the projections is the topology whose subbase is the collection $\left{\pi_\alpha^{-1}\left(U_\alpha\right): \alpha \in I\right.$, $\left.U_\alpha \in \mathcal{T}\alpha\right}$. We now formalize the above motivation to define the product topology $$\mathfrak{\Im}=\left{\pi\alpha^{-1}\left(U_\alpha\right): \alpha \in I, U_\alpha \in \mathcal{J}_\alpha\right}$$
Since $\cup{S: S \in \mathbb{S}}=X$, theorem 5.2.3 applies, and the following definition is meaningful.

# 数学分析代考

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

Days
Hours
Minutes
Seconds

# 15% OFF

## On All Tickets

Don’t hesitate and buy tickets today – All tickets are at a special price until 15.08.2021. Hope to see you there :)