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# 数学代写|数学物理方法代写Mathematical Methods代考|MATH0056 The Heine-Borel theorem

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## 数学代写|数学物理方法代写Mathematical Methods代考|The Heine-Borel theorem

1.0621. The Heine-Borel theorem. If every point of a closed interval $(a, b)$ is within some interval I of a family $F$, then there is a finite subfamily of $F$ such that every point of $(a, b)$ is within at least one interval of the subfamily. We say that $I$ covers $(c, d)$ if every point of $(c, d)$ is an interior point of $I$ (i.e. not an end-point).

There may be an interval $I$ belonging to $F$ that covers the whole of $(a, b)$. If so there is nothing to prove. If not, bisect $(a, b)$. There may be a pair of intervals $I_{1}, I_{2}$ such that every point of $\left(a, \frac{1}{2} a+\frac{1}{2} b\right)$ is interior to $I_{1}$ and every point of $\left(\frac{1}{2} a+\frac{1}{2} b, b\right)$ to $I_{2}$. If either half is not included in an interval $I$, bisect that half. We say that in a finite number of steps we shall arrive at a stage where every portion of $(a, b)$ lies within at least one interval $I$. For if not, the successive bisection of intervals will give a sequence of intervals, each part of the preceding one, and each half the length of the preceding one, and none of them included in an $I$. Such a sequence forms a nest of intervals and identifies a number $x_{0}$ common to all its members. But by hypothesis $x_{0}$ is interior to an $I$, say $I_{0}$, and hence there is a positive $\delta$ such that all points of $\left(x_{0}-\delta, x_{0}+\delta\right)$ are in $I_{0}$. Therefore all intervals of the nest whose lengths are less than $\delta$ are included in $I_{0}$, and we have a contradiction. Hence every division of $(a, b)$ is wholly interior to some $I$. Taking for each division an $I$ that includes it we have the theorem.

A slight modification is often made where an end-point, say $a$, is an end-point of an interval of the family, say $I_{a}$, closed at $a ; I_{a}$ is still supposed of non-zero length $\delta_{a}$. Then $a$ is interior to the interval $J_{a}\left(a-\frac{1}{2} \delta_{a}, a+\frac{1}{2} \delta_{a}\right)$, and the argument applies to the set of intervals $J$, where $J$ is the same as $I$ except that $I_{a}$ is replaced by $J_{a}$; every point of $(a, b)$ is an interior point of at least one $J$. But then the theorem follows with the modification that $a$ may be an end-point of $I_{a}$ or $b$ of $I_{b}$ provided that $I_{a}$ has $a$ as a member and $I_{b}$ has $b$.

The theorem gives the Bolzano-Weierstrass theorem (1-034) as a special case. If possible, let $(a, b)$ contain no limit-point of the set. Then every point of $(a, b)$ is in an interval $I$ containing not more than one member of the set. Hence $(a, b)$ can be covered by a finite set of such intervals $I$ and therefore contains only a finite number of members of the set of points considered, contrary to hypothesis.

In the argument as we have stated it the only intervals bisected at each stage are those not already covered by an $I$. We could, however, equally well bisect all the intervals. For if $I$ covers $(c, d)$ it covers both halves of it. Hence $(a, b)$ in the conditions stated can be divided into a finite set of equal intervals each covered by an $I$.

## 数学代写|数学物理方法代写Mathematical Methods代考|The modified Heine-Borel theorem

1.0622. The modified Heine-Borel theorem. In the Heine-Borel theorem the intervals $I$ may be specified by any rule so long as each is of non-zero length and every point of $(a, b)$ is an interior point of at least one of them (except that $a$ and $b$ may be end-points). Sometimes, however, a further restriction is made, according to which each point $x$ of $(a, b)$ specifies an $I_{x}$, of which $x$ is an interior point. Then the following theorem holds. Suppose that every point $x$ of $a \leqslant x \leqslant b$ is within an interval $I_{x}\left(x-\delta_{x}, x+\eta_{x}\right)$, where $\delta_{x}>0$, $\eta_{x}>0$, except that $I_{a}$ may be $a \leqslant x<a+\eta_{a}$ and $I_{b}$ may be $b-\delta_{b}<x \leqslant b$; then $(a, b)$ may be divided into a finite set of intervals such that each interval is part of the $I_{x}$ defined for some point of that interval. The proof is by successive bisection as before. Assuming the theorem false, we establish the existence of a nest of intervals converging to some $x_{0}$, such that none is part of $I_{x}$ for any $x$ within that interval; but all of them less than a certain length are parts of $I_{x_{0}}$, and contain $x_{0}$, and we have a contradiction. In this case, however, it does not follow that $(a, b)$ can be divided into equal intervals with the required property. If $(c, d)$ is covered by $I_{x}$, where $x$ is in $(c, d), x$ can be interior to only one half of $(c, d)$; then the other half is not necessarily covered by an $I_{v}$, when $y$ is now restricted to be in that half.

An important application is to differentiable functions. Let $f(x)$ be differentiable at all points of $a \leqslant x \leqslant b$; this says that for any $\omega$, for any $x$ in $(a, b)$, there is a positive $\delta(\omega, x)$ such that for $|h|<\delta \quad\left|f(x+h)-f(x)-h f^{\prime}(x)\right|<\omega|h|$ (1)
Then $(a, b)$ can be divided into a finite set of intervals $\left(x_{r}, x_{r+1}\right)$ such that for all $x$ of $\left(x_{r}, x_{r+1}\right)$
$$\left|f(x)-f\left(\xi_{r}\right)-\left(x-\xi_{r}\right) f^{\prime}\left(\xi_{r}\right)\right|<\omega\left|x-\xi_{r}\right| \text {, }$$
$\xi_{r}$ itself being a point of $\left(x_{r}, x_{r+1}\right)$.
For any fixed $\eta$ (1) remains true if all $\delta(\omega, x)$ are restricted to be $\leqslant \eta$. Then all $x_{r+1}-x_{r}$ will be $\leqslant 2 \eta$.

Heine* proved that a continuous function is uniformly continuous (1.071) by what was essentially a method of Dedekind section, capable of being used to prove the general Heine-Borel theorem and so used in Lebesgue’s proof. The specific use of overlapping intervals is due to Borel $\dagger$, the form of the Heine-Borel theorem given here to W.H.Young.f The bisection method was used by Bolzano; Goursat (see 11.043) used it in an important simplification of the conditions for Cauchy’s theorem, in which he recognized the effect of the restriction when each section is required to contain a point $x$ with which the $I_{x}$ covering that section is associated. He did not, however, give the general form of the modified theorem or comment on the possibility of proving the main theorem by the same method. This was first done by H. F. Baker in a note reported in title only. $\S$

# 数学物理方法代写

## 数学代写数学物理方法代写Mathematical Methods代想| The Heine-Borel theorem

1.0621. 海涅-博雷尔定理如果闭合区间的每个点 $(a, b)$ 在一个家庭的某个间隔内 $\mid F$ ，则有一个有限的亚科 $F$ 使得每个点 $(a, b)$ 至少 在亚科的一个区间内。我们说 $I$ 涵盖 $(c, d)$ 如果每个点 $(c, d)$ 是内部点 $I$ (即不是终点) 。

## 数学代写|数学物理方法代写Mathematical Methods代想| The modified HeineBorel theorem

1.0622. 修正的海涅-博雷尔定理.在海涅-博雷尔定理中，区间 $I$ 可以由任何规则指定，只要每个规则的长度都为非零，并且每个点 $(a, b)$ 是其中至少一个的内部点 (除了 $a$ 和 $b$ 可能是端点) 。但是，有时还会进行进一步的限制，根据该限制，每个点 $x$ 之 $(a, b)$ 指 定 $I_{x}$ ，其中 $x$ 是一个内部点。那么下面的定理成立。假设每个点 $x$ 之 $a \leqslant x \leqslant b$ 在某个时间间隔内 $I_{x}\left(x-\delta_{x}, x+\eta_{x}\right)$ 哪里 $\delta_{x}>0, \eta_{x}>0$ ，除了 $I_{a}$ 可能 $a \leqslant x<a+\eta_{a}$ 和 $I_{b}$ 可能 $b-\delta_{b}<x \leqslant b$;然后 $(a, b)$ 可以划分为一组有限的区间，使得每个区 间都是 $I_{x}$ 为该间隔的某个点定义。证明是像以前一样通过连续的对等分。假设定理为假，我们确定存在一个收敀于某个区间的緖 穴 $x_{0}$ ，使得没有一个是 $I_{x}$ 对于任何 $x$ 在该间隔内; 但所有小于一定长度的都是 $I_{x_{0} ｝ \text { ，并包含 } x_{0} \text { ，我们有一个矛盾。但是，在这种情 }$ 况下，这并不意味着 $(a, b)$ 可以使用所需属性划分为相等的间隔。如果 $(c, d)$ 涵盖 $I_{x}$ 哪里 $x$ 位于 $(c, d), x$ 可以内部只有一半 $(c, d)$ ； 那么另一半不一定被 $I_{v}$ 什么时候 $y$ 现在被限制在那一半。

$$\left|f(x)-f\left(\xi_{r}\right)-\left(x-\xi_{r}\right) f^{\prime}\left(\xi_{r}\right)\right|<\omega\left|x-\xi_{r}\right|,$$
$\xi_{r}$ 本身就是一个点 $\left(x_{r}, x_{r+1}\right)$.

Heine* 通过本质上是戴德金截面的方法证明了连续函数是均匀连续的（1.071)，能够用于证明一般的 Heine-Borel 定理，因此 在勒贝格的证明中也使用了这种函数。重殞间隔的具体使用是由于Borel†，这里给出给W.H.Young.f的Heine-Borel定理的形式， Bolzano使用了对等分法;Goursat (见11.043) 在柯西定理条件的重要简化中使用它，其中他认识到当每个部分都需要包含一个 点时，限制的效果。 $x$ 与哪个 $I_{x}$ 覆盖该部分是关联的。然而，他没有给出修改定理的一般形式，也没有评论用同样的方法证明主要 定理的可能性。这是H. F. Baker首先在标题中报道的注释中完成的。

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