MY-ASSIGNMENTEXPERT™可以为您提供sydney MATH5340 Topology代数拓扑的代写代考和辅导服务！
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）
(O) Give an example of subsets $A, B \subset \mathbb{R}$ such that
$$
A \cap B=\emptyset, \quad \bar{A} \cap B \neq \emptyset, \quad A \cap \bar{B} \neq \emptyset .
$$
For example, $A=[0,1[$ and $B=\mathbb{R}-A$.
(৩) Let $A, B$ be subsets in a topological space. Prove $\overline{A \cup B}=\bar{A} \cup \bar{B}$.
From $A \subset A \cup B$ follows $\bar{A} \subset \overline{A \cup B}$; similarly $\bar{B} \subset \overline{A \cup B}$ implies $\bar{A} \cup \bar{B} \subset$ $\overline{A \cup B}$. On the other hand $\bar{A} \cup \bar{B}$ is closed and contains $A \cup B$, so $\overline{A \cup B} \subset \bar{A} \cup \bar{B}$.
(\bigcirc)$ Let $A$ be a dense subset of a topological space $X$. Prove $\overline{U \cap A}=\bar{U}$ for any open set $U \subset X$.
As $A \cap U \subset U$, we have $\overline{U \cap A} \subset \bar{U}$; therefore it suffices to prove that $\bar{U} \subset$ $\overline{U \cap A}$, or equivalently, that the closed set $\overline{U \cap A}$ contains $U$. But the set $V=$ $U \cap(X-\overline{U \cap A})$ is open, and since $A \cap V=\emptyset$ by the density of $A$ we obtain $V=\emptyset$.
(O) Let $X$ be a set and suppose we are given, for any $x \in X$, a family $\mathcal{I}(x)$ of subsets in $X$ so that the following five conditions hold:
- $X \in \mathcal{I}(x)$ for every point $x \in X$;
- $x \in U$ for any $U \in \mathcal{I}(x)$;
- if $U \in \mathcal{I}(x)$ and $U \subset V$, then $V \in \mathcal{I}(x)$;
- if $U, V \in \mathcal{I}(x)$, then $U \cap V \in \mathcal{I}(x)$;
- if $U \in \mathcal{I}(x)$, there is a subset $V \subset U$ such that $x \in V$ and $V \in \mathcal{I}(y)$ for any $y \in V$.
Prove that there exists a unique topology on $X$ for which $\mathcal{I}(x)$ is the family of neighbourhoods at $x$, for any $x \in X$.
Define a topology $\mathcal{T}$ on $X$ by declaring open the subsets that are neighbourhoods of all of their points, i.e. $A \in \mathcal{T}$ iff $A \in \mathcal{I}(x)$ for any $x \in A$. Let us check that in this way (A1), (A2), (A3) from Definition 3.1 hold true.
The empty set, having no elements, is a neighbourhood for all its points, while condition 1 implies that $X$ is open. Let $\left{A_i \mid i \in I\right}$ be a family of open sets and call $A=\cup\left{A_i \mid i \in I\right}$ their union. If $x \in A$ then $x \in A_j$ for some $j$, so $A_j$ is a neighbourhood of $x$ and by condition 3 it follows that $A$ is also a neighbourhood of $x$. If $A$ and $B$ are open sets and $x \in A \cap B$, then $A, B \in \mathcal{I}(x)$ and $A \cap B \in \mathcal{I}(x)$ by condition 4.
Now that we know $\mathcal{T}$ is a topology let’s denote, just for the moment, by $J(x)$ the family of neighbourhoods of $x$ in $\mathcal{T}$. We claim $\mathcal{I}(x)=J(x)$. If $U \in J(x)$ there is an open set $A$ such that $x \in A$ and $A \subset U$. By definition $A \in \mathcal{I}(x)$, so condition 3 implies $U \in \mathcal{I}(x)$. Vice versa, take $U \in \mathcal{I}(x)$ and $V$ as in condition 5 . Then $V \in \mathcal{T}$ and so $U \in J(x)$.
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