# 度量空间和基本拓扑学代写|METRIC SPACE AND BASIC TOPOLOGY MATHS4077 University of Glasgow Assignment

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## Instructions:

In a course on topology, you can expect to study the fundamental concepts of point-set topology, including sets, functions, continuity, convergence, compactness, and connectedness. You will learn how to use these concepts to prove theorems and solve problems in a variety of mathematical contexts.

The course will likely start with a review of basic set theory and logic, and then move on to the definition and properties of metric spaces. You will learn about various topological concepts and constructions, such as open sets, closed sets, neighborhoods, interior, closure, and boundary. You will study different types of convergence, such as pointwise, uniform, and Hausdorff convergence, and learn how to use these concepts to prove continuity and other results.

Later on, you may also study more advanced topics in topology, such as homotopy theory, algebraic topology, and differential topology. These areas of study involve the use of topological concepts and tools to solve more sophisticated problems in mathematics and other fields.

Overall, a course in topology will provide you with a deep understanding of the fundamental concepts and methods of topology, which are essential for further study in mathematics and other areas of science and engineering.

Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x)=k x+b$ for $0<k<1$ and $b \in \mathbb{R}$. Show that $f$ is a contraction, find the fixed point of $f$, and directly show the fixed point is unique.

To show that $f$ is a contraction, we need to show that there exists a constant $0 \leqslant L < 1$ such that for all $x, y \in \mathbb{R}$, we have $|f(x) – f(y)| \leqslant L|x-y|$.

So let $x, y \in \mathbb{R}$ be arbitrary, and let $L = k$. Then we have

\begin{align*} |f(x) – f(y)| &= |kx + b – ky – b| \ &= |k(x-y)| \ &= k|x-y|. \end{align*}

Since $0<k<1$, we have $0\leqslant L<k<1$. Therefore, $f$ is a contraction.

To find the fixed point of $f$, we need to solve the equation $f(x) = x$. That is,

$$kx+b=x$$

which gives $x = \frac{b}{1-k}$. Therefore, the fixed point of $f$ is $\frac{b}{1-k}$.

To show that the fixed point is unique, we need to show that if $x_1$ and $x_2$ are both fixed points of $f$, then $x_1 = x_2$.

So let $x_1$ and $x_2$ be fixed points of $f$. Then we have

\begin{align*} f(x_1) &= x_1 \ f(x_2) &= x_2 \end{align*}

Subtracting these equations, we get

\begin{align*} f(x_1) – f(x_2) &= x_1 – x_2 \ kx_1 + b – kx_2 – b &= x_1 – x_2 \ k(x_1 – x_2) &= x_1 – x_2 \ \end{align*}

Since $k < 1$, we can divide both sides by $k(x_1 – x_2)$ to obtain $1 < \frac{1}{k}$, which means $x_1 = x_2$. Therefore, the fixed point of $f$ is unique.

In class, we have defined a set $A \subset X$ to be closed if its complement is an open set in $X$. There is another useful definition of a closed set however. Show that $A \subset X$ is closed if and only if every convergent sequence in $A$ converges in $A$. In other words, if $\left\{x_n\right\}$ is a convergent sequence in $A$ such that $x_n \rightarrow x$, then $x \in A$.

To prove that $A \subset X$ is closed if and only if every convergent sequence in $A$ converges in $A$, we need to show two things:

$(\Rightarrow)$ If $A$ is closed, then every convergent sequence in $A$ converges in $A$. $(\Leftarrow)$ If every convergent sequence in $A$ converges in $A$, then $A$ is closed.

$(\Rightarrow)$ Suppose that $A$ is closed, and let $\left{x_n\right}$ be a convergent sequence in $A$ that converges to some $x \in X$. We want to show that $x \in A$. Since $\left{x_n\right}$ converges to $x$, we know that for any open set $U$ containing $x$, there exists an $N$ such that $x_n \in U$ for all $n \geq N$. In particular, for the open set $U = X \setminus A$, we have $x_n \notin U$ for all $n \geq N$, since $\left{x_n\right}$ is a sequence in $A$. Therefore, $x \notin U$, which implies that $x \in A$ (since $U = X \setminus A$ is closed). Thus, every convergent sequence in $A$ converges to a point in $A$, as required.

$(\Leftarrow)$ Suppose that every convergent sequence in $A$ converges in $A$, and let $U = X \setminus A$. We want to show that $U$ is open. Let $x \in U$ be arbitrary. Since $x \notin A$, there exists a sequence $\left{x_n\right}$ in $A$ that converges to $x$. But by assumption, this means that $x \in A$, which is a contradiction. Therefore, $U$ contains no points of $A$, and is therefore disjoint from $A$. Thus, $U$ is open, and $A$ is closed.

Therefore, we have shown that $A$ is closed if and only if every convergent sequence in $A$ converges in $A$.

Let $|\cdot|$ be a norm on a vector space $V$, and let $d(x, y)=|x-y|$ for all $x, y \in V$. Show the following three properties:
(a) $d(\lambda x, \lambda y)=|\lambda| d(x, y)$ for all $\lambda \in \mathbb{R}$, and for all $x, y \in V$.

To show that $d(\lambda x, \lambda y) = |\lambda| d(x, y)$ for all $\lambda \in \mathbb{R}$, and for all $x, y \in V$, we need to use the properties of the norm $|\cdot|$.

First, note that by the definition of $d$, we have:

$d(\lambda x, \lambda y)=|\lambda x-\lambda y|$

Using the properties of the norm, we can manipulate this expression as follows:

\begin{align*} |\lambda x – \lambda y| &= |\lambda (x – y)| \ &= |\lambda| |x – y| \ &= |\lambda| d(x, y). \end{align*}

Therefore, we have shown that $d(\lambda x, \lambda y) = |\lambda| d(x, y)$ for all $\lambda \in \mathbb{R}$, and for all $x, y \in V$. This property is known as homogeneity or scaling property of the norm.