## 数学代写|勒贝格积分代写Lebesgue Integration代考|Math720

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## 数学代写|勒贝格积分代写Lebesgue Integration代考|Accommodating Algebra

The real number line entails more than distance. We want to assign values to its points. We choose a point that will represent the origin, 0 , and then pick a second point, label it “1,” and use the distance between the origin and 1 as our basic linear unit. We can locate points whose distances from the origin correspond to integers and rational numbers. We can even locate points that correspond to irrational lengths such as $\sqrt{2}$, the length of the diagonal of a unit square. Taking mirror images across the origin, we locate the negatives of these numbers. We have imposed a system of a discrete set of objects – integers, rational numbers, algebraic numbers – onto our continuum of distances. But this system does not account for all points in $\mathbb{R}$. In 1844, Joseph Liouville proved that there are points on the real line that are not algebraic.

It is a very small and self-evident step to believe that every point on the continuum of the real line corresponds to a number, but this step carries enormous repercussions, for from now on we will be using the geometric notion of distance to inform our concept of number that, until now, had been restricted to quantities arising from algebraic constructions.

Although some mathematicians resisted expanding the notion of number beyond algebraic numbers, most recognized the need to do so. Foremost among them were Richard Dedekind and Karl Weierstrass.

Beginning in the academic year 1857-1858 and then every 2 years until the $1880 \mathrm{~s}$, Weierstrass lectured on analysis at the University of Berlin. In these lectures, he developed and expounded many of the basic principles of analysis. His students would work through these ideas, refine them, and eventually publish them. It was in these lectures that Weierstrass first explained what today we call the Bolzano-Weierstrass theorem. His proof rested on the nested interval principle. The shared attribution with Bernhard Bolzano arises from Weierstrass’s acknowledgment of his indebtedness to Bolzano’s 1817 proof that the convergence of every Cauchy sequence implies that every bounded, increasing sequence has a limit.

## 数学代写|勒贝格积分代写Lebesgue Integration代考|Harnack’s Mistake

We take the ordering of the rational points in $[0,1]$ given in $(3.2)$ and call them $\left(a_1=\right.$ $\left.0, a_2=1, a_3=1 / 2, a_4=1 / 3, \ldots\right)$. We choose any positive $\epsilon$ and let $I_k$ be the open interval of length $\epsilon / 2^k$ that is centered at $a_k: I_k=\left(a_k-\epsilon / 2^{k+1}, a_k+\epsilon / 2^{k+1}\right)$. Does the union of these intervals contain all points in $[0,1]$ ? In other words, can we put the closed interval $[0,1]$ inside a countable union of open intervals whose lengths add up to $\epsilon$ ? This was a problem first posed by Axel Harnack in 1885 . He convinced himself that the answer is “yes.”

Axel Harnack (1851-1888) was the younger twin brother of the German theologian Adolf von Harnack. Axel earned his doctorate at Erlangen-Nürnberg University in 1875 , working under the direction of Felix Klein. He is best known for his work in harmonic analysis and the theory of algebraic curves.

In essence, what Harnack did was to ask himself, “What is the complement of a countable union of intervals?” He believed that it must also be a countable union of intervals. Think about this. The intervals might be open or closed or half-open/halfclosed, and a closed interval might be a single point. It is certainly true that the complement of any finite union of intervals is a finite union of intervals. It is not obvious that the same would not be true for countable unions. But if Harnack was right, then the complement of $\bigcup I_k$ is a countable union of intervals. The intervals in the complement must be single points, otherwise they would contain rational numbers between 0 and 1 . We now put each of these countably many points inside intervals whose lengths add up to $\epsilon$, and we now have all of $[0,1]$ contained within a union of countably many intervals whose lengths add up to $2 \epsilon$.

In fact, Harnack’s basic premise, that the complement of a countable union of intervals is a countable union of intervals, is wrong. The complement can be an uncountable union. Georg Cantor and others understood this. The flaw in Harnack’s reasoning underscores some of the complexity of the real number line as a set of numbers, and we shall treat it in full detail in the next section. But that does not prove that his answer was wrong. This was accomplished by Émile Borel in 1895.

# 勒贝格积分代考

## 有限元方法代写

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## MATLAB代写

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