# 数学代写|拓扑学代写Topology代考|COMPACT SPACES

## 数学代写|拓扑学代写Topology代考|COMPACT SPACES

Let $X$ be a topological space. A class $\left{G_i\right}$ of open subsets of $X$ is said to be an open cover of $X$ if each point in $X$ belongs to at least one $G_i$, that is, if $\cup_i G_i=X$. A subclass of an open cover which is itself an open cover is called a subcover. A compact space is a topological space in which every open cover has a finite subcover. A compact subspace of a topological space is a subspace which is compact as a topological space in its own right. We begin by proving two simple but widely used theorems.
Theorem A. Any closed subspace of a compact space is compact. proof. Let $Y$ be a closed subspace of a compact space $X$, and let $\left{G_i\right}$ be an open cover of $Y$. Each $G_i$, being open in the relative topology on $Y$, is the intersection with $Y$ of an open subset $H_i$ of $X$. Since $Y$ is closed, the class composed of $Y^{\prime}$ and all the $H_i{ }^{\prime}$ ‘s is an open cover of $X$, and since $X$ is compact, this open cover has a finite subcover. If $Y^{\prime}$ occurs in this subcover, we discard it. What remains is a finite class of $H_2$ ‘s whose union contains $X$. Our conclusion that $Y$ is compact now follows from the fact that the corresponding $G_i$ ‘s form a finite subcover of the original open cover of $Y$.
Theorem B. Any continuous image of a compact space is compact. Proof. Let $f: X \rightarrow Y$ be a continuous mapping of a compact space $X$ into an arbitrary topological space $Y$. We must show that $f(X)$ is a compact subspace of $Y$. Let $\left{G_i\right}$ be an open cover of $f(X)$. As in the above proof, each $G_{\Delta}$ is the intersection with $f(X)$ of an open subset $H_i$ of $Y$. It is clear that $\left{f^{-1}\left(H_i\right)\right}$ is an open cover of $X$, and by the compactness of $X$ it has a finite subcover. The union of the finite class of $H_i$ ‘s of which these are the inverse images clearly contains $f(X)$, so the class of corresponding $G_i$ ‘s is a finite subcover of the original open cover of $f(X)$, and $f(X)$ is compact.

It is sometimes quite difficult to prove that a given topological space is compact by appealing directly to the definition. The following theorems give several equivalent forms of compactness which are of ten easier to apply.

## 数学代写|拓扑学代写Topology代考|PRODUCTS OF SPACES

There are two main techniques for making new topological spaces out of old ones. The first of these, and the simplest, is to form subspaces of some given space. The second is to multiply together a number of given spaces. Our purpose in this section is to describe the way in which the latter process is carried out.

In Sec. 4 we defined what is meant by the product $P_i X_i$ of an arbitrary non-empty class of sets. We also defined the projection $p_i$ of this product onto its $i$ th coordinate set $X_i$. The reader should make certain that these concepts are firmly in mind. If each coordinate set is a topological space, then there is a standard method of defining a topology on the product. It is difficult to exaggerate the importance of this definition, and we examine it with great care in the following discussion.
Let us begin by recalling the discussion in Sec. 18 of open rectangles and open strips in the Euclidean plane $R^2$. We observed there that the open rectangles form an open base for the topology of $R^2$, and also that the open strips form an open subbase for this topology whose generated open base consists of all open rectangles, all open strips, the empty set, and the full space. The topology of the Euclidean plane is of course defined in terms of a metric. If we wish, however, we can ignore this fact and regard the topology of $R^2$ as generated in the sense of Theorem 18-D by the class of all open strips. This situation provides the motivation for the more general ideas we now develop.

Let $X_1$ and $X_2$ be topological spaces, and form the product $X=X_1 \times X_2$ of the two sets $X_1$ and $X_2$. Consider the class $\mathbf{S}$ of all subsets of $X$ of the form $G_1 \times X_2$ and $X_1 \times G_2$, where $G_1$ and $G_2$ are open subsets of $X_1$ and $X_2$, respectively. The topology on $X$ generated by this class in the sense of Theorem 18-D is called the product topology on $X$.

# 拓扑学代考

## 数学代写|拓扑学代写Topology代考|PRODUCTS OF SPACES

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: