数学代写|Math432 Topology

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Math432课程简介

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Prerequisite: Minimum grade of C- in MATH410. Metric spaces, topological spaces, connectedness, compactness including Heine-Borel and Bolzano-Weierstrass theorems, Cantor sets, continuous maps and homeomorphisms, fundamental group homotopy, covering spaces, the fundamental theorem of algebra, Brouwer fixed point theorem, surfaces e.g., Euler characteristic, the index of a vector field, hairy sphere theorem, elements of combinatorial topology (graphs and trees, planarity, coloring problems.Credits3

Prerequisites 

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Math432 Topology HELP（EXAM HELP， ONLINE TUTOR）

Without using the Axiom of Choice, it is not possible to construct a choice function for the collection $\mathcal{A}$ of nonempty subsets of $\mathbb{Z}_{+}$.

To see this, consider the collection $\mathcal{A}’$ of nonempty subsets of $\mathbb{Z}_{+}$ that do not contain the number $1$. Then for any $A\in\mathcal{A}’$, we can define a choice function $f$ by setting $f(A)=\min A$. However, $\mathcal{A}’$ is nonempty and has no choice function, because if we take the union of any two sets in $\mathcal{A}’$, the resulting set will also be in $\mathcal{A}’$ and will not have a minimum element.

This shows that the collection $\mathcal{A}$ of nonempty subsets of $\mathbb{Z}_{+}$ cannot have a choice function without the Axiom of Choice.

It is possible to construct a choice function for the collection $\mathcal{B}$ of nonempty subsets of $\mathbb{Q}$ without using the Axiom of Choice.

Let $A\in\mathcal{B}$ be a nonempty subset of $\mathbb{Q}$. We can define a choice function $f$ for $\mathcal{B}$ as follows:

1. Choose an element $q_0\in A$.
2. For each positive integer $n$, let $q_n$ be the element of $A$ with the smallest denominator among those with denominator at most $n$ (if there are multiple such elements, choose the one with the smallest numerator).
3. Define $f(A)=\lim_{n\to\infty}q_n$.

Since $A$ is nonempty, we can always choose an element $q_0\in A$. The sequence $(q_n)$ is well-defined and nonempty because $A$ is nonempty, and it is Cauchy because the denominators are bounded. Therefore, the limit $\lim_{n\to\infty}q_n$ exists in $\mathbb{Q}$.

It remains to show that $f(A)$ is an element of $A$. Suppose not; then there exists an element $a\in A$ such that $a<f(A)$. Since $f(A)$ is the limit of $(q_n)$, there exists a positive integer $N$ such that $q_n>f(A)-\frac{a-f(A)}{2}$ for all $n\geq N$. But this implies that $a>f(A)-\frac{a-f(A)}{2}$, or equivalently, $a>\frac{f(A)}{2}+\frac{a}{2}$, which is a contradiction. Therefore, $f(A)\in A$, and $f$ is a choice function for $\mathcal{B}$.

Note that the above construction relies on the fact that $\mathbb{Q}$ is an ordered field, which allows us to use the idea of a “smallest denominator.” This technique would not work for collections of subsets of more general sets.

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