# 英国代考|MATH0003 Analysis 1数学分析1

Proof. Indeed, if (a) holds, then there exists $\varepsilon>0$ such that for every $\delta>0$, we can find the numbers $x, y \in(c, c+\delta)$ such that
$$|f(x)-f(c)| \geq \sqrt{\varepsilon} \text { and } u(y)-u(c) \geq \sqrt{\varepsilon} .$$
Let $P=\left{x_{0}, \dots, x_{n}\right}$ be any partition of $[a, b]$. Then $c \in\left[x_{k}, x_{k+1}\right)$ for some $k=$ $0, \ldots, n-1$. Let $x \in\left(c, x_{k+1}\right)$ be so that $|f(x)-f(c)| \geq \sqrt{\varepsilon}$. Then there exists $y \in(c, x)$ such that $u(y)-u(c) \geq \sqrt{\varepsilon}$. Since $u$ is increasing, $u(x)-u(c) \geq u(y)-$ $u(c) \geq \sqrt{\varepsilon}$. Now consider the refinement
$$Q=\left{x_{0}, \ldots, x_{k-1}, x_{k}, c, x, x_{k+1}, \ldots, x_{n}\right}$$
of $\mathcal{P}$. Note that if $c \in \mathcal{P}$, then $c$ and $x_{k}$ are same numbers. Then
$$S^{}(f, u, \mathcal{Q})-S_{}(f, u, \mathcal{Q}) \geq|f(x)-f(c)|(u(x)-u(c)) \geq \varepsilon .$$
This implies $S^{}(f, u, \mathcal{P})-S_{}(f, u, \mathcal{P}) \geq \varepsilon$ since $\mathcal{Q}$ is a refinement of $\mathcal{P}$. From the arbitrariness of $\mathcal{P}$, we conclude that $S^{}(f, u)-S_{}(f, u) \geq \varepsilon$, that is, $(f, u) \notin R S(a, b)$. The arguments are similar if (b) holds.

Step 6: $a \in B^{c}$ is true (follows from Step 5).
Step 7: $a \in A^{c} \cap B^{c}$ is true (conclusion, follows from Steps 4 and 6).
To get a proof that $Q \Rightarrow P$ is true, we verify whether or not it can be obtained by conversion in the order of the steps in the preceding proof. This is an easy way to get a proof of the truth value of a converse proposition that, unfortunately, is not always a successful way. So, we obtain the following (direct) proof by conversion that $Q \Rightarrow P$ is true:
Step 1: $a \in A^{c} \cap B^{c}$ is true (condition, it is assumed).
Step 2: $a \in B^{c}$ is true (follows from Step 1).
Step 3: $a \notin B$ is true (follows from Step 2).
Step 4: $a \in A^{c}$ is true (follows from Step 1).
Step 5: $a \notin A$ is true (follows from Step 4).
Step 6: $a \notin A \cup B$ is true (follows from Steps 3 and 5).
Step 7: $a \in(A \cup B)^{c}$ is true (conclusion, follows from Step 6).

myassignments-hel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。

physics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: