# 微积分网课代修|积分学代写Integral Calculus代考|MATH122 Cousin covering lemma

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## 微积分作业代写calclulus代考|Cousin covering lemma

LEMMA $2.4$ (Cousin). Let $\beta$ be a full cover. Then $\beta$ contains a partition of every compact interval $[a, b]$, i.e., there is a subset $\pi$ of $\beta$,
$$\pi=\left{\left(\left[a_{i}, b_{i}\right], \xi_{i}\right): i=1,2,3, \ldots, n\right}$$
so that the intervals $\left[a_{i}, b_{i}\right]$ are nonoverlapping and combine to form the whole interval $[a, b]$.
PROOF IN SECTION 7.5.1.
COROLLARY 2.5. Let $\beta$ be a Cousin cover of a compact interval $[a, b]$. Then $\beta$ contains a partition of every compact subinterval of $[a, b]$.
PROOF IN SECTION 7.5.2.

## 微积分作业代写calclulus代考|An application of the Cousin lemma

The Cousin lemma offers us a technique that can be used to prove Lemma $1.5$ that we just skipped over. This lemma follows from the following statement which we now prove, using the Cousin covering lemma:
Suppose that $F$ is a continuous function. Suppose that there is a sequence of points $e_{1}, e_{2}, e_{3}, \ldots$ of points and that $F^{\prime}(x)=0$ for all $x$ except possibly at the points $e_{1}, e_{2}, e_{3}, \ldots$ Then $F$ is constant.
The next chapter is devoted to an elaborate study of this lemma, in a more general exposition. It is, nonetheless, worth working through the details here as a preliminary to the studying the collection of definitions and techniques to be used there.

Proof. Fix an interval $[a, b]$. The proof is obtained by showing, for any $\epsilon>0$, that
$$|F(b)-F(a)|<\epsilon .$$ This can only be true for all $\epsilon>0$ if $F(b)=F(a)$. This shows that $F$ is in fact constant.

We use a covering argument. We need to construct a full cover $\beta$ using a different construction at the points where $F^{\prime}(x)=0$ and at the points $e_{1}, e_{2}$, … where the derivative need not be zero. There are infinitely many steps so this value of $\epsilon$ will need to be split into infinitely many small pieces by the following simple identity:
$$\epsilon=\frac{\epsilon}{2}+\frac{\epsilon}{4}+\frac{\epsilon}{8}+\frac{\epsilon}{16}+\ldots$$
The first step in the covering argument takes advantage of the fact that
$$F^{\prime}(x)=0$$
at points other than $e_{1}, e_{2}, \ldots$ Write
$$\eta=\frac{\epsilon}{2(b-a)}$$
and
$$\beta_{1}={([c, d], x): a<c \leq x \leq d<b \text { and }|F(d)-F(c)|<\eta(d-c)}$$

## 微积分作业代写calclulus代考|Cousin covering lemma

$\backslash$ pi=\left } { \backslash \text { left(\left[a_{i}, b_{i}\right } ] , \backslash x i _ { – } { i } \backslash \text { right } ) : i = 1 , 2 , 3 , \backslash \text { ldots, } n \backslash \text { right } }

## 微积分作业代写calclulus代考|An application of the Cousin lemma

$$|F(b)-F(a)|<\epsilon .$$ 这只能适用于所有人 $\epsilon>0$ 如果 $F(b)=F(a)$. 这表明 $F$ 实际上是恒定的。

$$\epsilon=\frac{\epsilon}{2}+\frac{\epsilon}{4}+\frac{\epsilon}{8}+\frac{\epsilon}{16}+\ldots$$

$$F^{\prime}(x)=0$$

$$\eta=\frac{\epsilon}{2(b-a)}$$

$$\beta_{1}=([c, d], x): a<c \leq x \leq d<b \text { and }|F(d)-F(c)|<\eta(d-c)$$