数学代写|MATH5340 Topology

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这是悉尼大学拓扑学课程的成功案列。

MATH5340课程简介

Topology is the mathematical theory of the “shape of spaces”. It gives a flexible framework in which the fabric of space is like rubber and thus enables the study of the general shape of a space. The spaces often arise indirectly: as the solution space of a set of equations; as the parameter space for a family of objects; as a point cloud from a data set; and so on. This leads to strong interactions between topology and a plethora of mathematical and scientific areas. The love of the study and use of topology is far reaching, including the use of topological techniques in the phases of matter and transition which received the 2016 Nobel Prize in Physics. This unit introduces you to a selection of topics in pure or applied topology. Topology receives strength from its areas of applications and imparts insights in return. A wide spectrum of methods is used, dividing topology into the areas of algebraic, computational, differential, geometric and set-theoretic topology. You will learn the methods, key results, and role in current mathematics of at least one of these areas, and gain an understanding of current research problems and open conjectures in the field.

Prerequisites 

At the completion of this unit, you should be able to:

• LO1. Demonstrate a coherent and advanced understanding of the key concepts of fundamental group, covering spaces, homology and cohomology.
• LO2. Apply the fundamental principles and results of algebraic topology to solve given problems.
• LO3. Distinguish and compare the properties of different types of topological spaces and maps between them.
• LO4. Formulate topological problems in terms of algebraic invariants and determine the appropriate framework to solve them.
• LO5. Communicate coherent mathematical arguments appropriately to student and expert audiences, both orally and through written work.
• LO6. Devise computational solutions to complex problems in algebraic topology.
• LO7. Compose correct proofs of unfamiliar general results in algebraic topology.

MATH5340 Algebraic topology HELP（EXAM HELP， ONLINE TUTOR）

If $X$ contains just one point there’s nothing to be proven. Assume then $X$ is finite, with at least two points. Take $x \in X$ and set
$$
r=\min {d(x, y) \mid y \in X, y \neq x} .
$$
Clearly $r>0$ and $B(x, r)={x}$, from which the topology on $X$ is discrete, and any subset is open and closed. Observe that there exists $y \in X$ such that $d(x, y)=r$, whence
$$
{x}=\overline{B(x, r)} \neq{y \in X \mid d(x, y) \leq r} .
$$

Both assertions are false. One counterexample is the discrete subspace ${1 / n \mid$ $n \in \mathbb{N}} \subset \mathbb{R}$

We want to prove that any discrete subspace $X \subset \mathbb{R}$ is countable. Given $x \in X$ choose a positive real $h(x)$ such that
$$
X \cap B(x, h(x))={x},
$$
and for any integer $n>0$ take the subset
$$
X_n={x \in X|| x \mid1 / n} .
$$
Since $X=\cup_n X_n$ it’s enough to show that every $X_n$ is finite. For any pair of distinct points $x, y \in X_n$ we have $|x-y| \geq \max (h(x), h(y)) \geq 1 / n$, so $X_n$ contains $2 n^2$ points at most.

3.61 Take four natural numbers $a, b, c, d$ such that $\operatorname{gcd}(a, b)=1$ and $\operatorname{gcd}$ $(c, d)=1$. The product $b d$ is coprime to any element in $N_{a, b} \cap N_{c, d}$ : in fact, if $n=a+k b=c+h d$ and $p$ is a prime that divides $b d$, then either $p$ divides $b$ and then $\operatorname{gcd}(p, n)=\operatorname{gcd}(p, a)=1$, or $p$ divides $d$ and so $\operatorname{gcd}(p, n)=\operatorname{gcd}(p, c)=1$. Hence
$$
\mathbb{N}=N_{1,1}, \quad N_{a, b} \cap N_{c, d}=\cup\left{N_{n, b d} \mid n \in N_{a, b} \cap N_{c, d}\right}
$$
showing that $\mathcal{B}$ is a basis of a topology $\mathcal{T}$.
Given distinct positive integers $n, m$, for any prime $p>\max (n, m)$ we have
$$
n \in N_{n, p}, \quad m \in N_{m, p}, \quad N_{n, p} \cap N_{m, p}=\emptyset,
$$
so that $\mathcal{T}$ is a Hausdorff topology.
Let $a, b$ be coprime, fix a multiple $h b$ of $b$ and let’s show that $h b \in N_{c, d}$ implies
$$
N_{a, b} \cap N_{c, d} \neq \emptyset .
$$
From $h b \in N_{c, d}$ follows that $b$ and $d$ don’t have common divisors, so we can find positive integers $t, s$ such that $t b-s d=c-a$. Then
$$
a+t b=c+s d \in N_{a, b} \cap N_{c, d} .
$$
Take $A, B$ open, non-empty; then there exist four natural numbers $a, b, c, d$ such that $\operatorname{gcd}(a, b)=\operatorname{gcd}(c, d)=1$ and
$$
N_{a, b} \subset A, \quad N_{c, d} \subset B .
$$
By what we saw above,
$$
b d \in \overline{N_{a, b}} \cap \overline{N_{c, d}} \subset \bar{A} \cap \bar{B} .
$$
In any metric space $(X, d)$ of cardinality larger than one we may always find two open non-empty sets with disjoint closures, namely
$$
\overline{B(x, r)} \subset{z \in X \mid d(x, z) \leq r},
$$
for $x \in X$ and $r$ a real positive number. Given distinct $x, y$ and a real number $r<d(x, y) / 2$, the triangle inequality gives
$$
\overline{B(x, r)} \cap \overline{B(y, r)}=\emptyset .
$$

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