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这是大峡谷州立大学拓扑学课程的代写成功案例。
MTH441课程简介
An introduction to the fundamental concepts of topology. The topology of the real number system and its generalizations to metric spaces and topological spaces. Topics include subspaces, neighborhood spaces, open and closed sets, interior and boundary of sets, continuity and homeomorphisms, connected and locally connected spaces, compact sets and spaces. Prerequisites: MTH 203, MTH 210, and MTH 204.
Credits: 3
Prerequisites
Topology is the mathematical study of the properties that are preserved through deformations, twistings, and stretchings of objects. Tearing, however, is not allowed. A circle is topologically equivalent to an ellipse (into which it can be deformed by stretching) and a sphere is equivalent to an ellipsoid. Similarly, the set of all possible positions of the hour hand of a clock is topologically equivalent to a circle (i.e., a one-dimensional closed curve with no intersections that can be embedded in two-dimensional space), the set of all possible positions of the hour and minute hands taken together is topologically equivalent to the surface of a torus (i.e., a two-dimensional a surface that can be embedded in three-dimensional space), and the set of all possible positions of the hour, minute, and second hands taken together are topologically equivalent to a three-dimensional object.
MTH441 Topology HELP（EXAM HELP， ONLINE TUTOR）
Let $f: X \rightarrow Y$ be a continuous mapping of a compact space $X$ onto a Hausdorff space $Y$. Prove that $f$ is a closed map, and hence a quotient map.
To show that $f$ is a closed map, we need to prove that the image of any closed subset $C$ of $X$ under $f$ is a closed subset of $Y$.
Let $C$ be a closed subset of $X$. Since $f$ is continuous, the image of $C$ under $f$, denoted by $f(C)$, is a closed subset of $Y$ if and only if its complement $Y \setminus f(C)$ is open in $Y$.
Let $y \in Y \setminus f(C)$. Since $f$ is onto, there exists $x_y \in X$ such that $f(x_y) = y$. Since $Y$ is Hausdorff, for each $x \in C$, we can find open neighborhoods $U_x$ of $y$ and $V_x$ of $f(x)$ such that $U_x \cap V_x = \varnothing$. Since $C$ is compact, there exist finitely many points $x_1, x_2, \ldots, x_n$ in $C$ such that $C \subseteq \bigcup_{i=1}^n V_{x_i}$. Then, the open set $U = \bigcap_{i=1}^n U_{x_i}$ is an open neighborhood of $y$ that does not intersect $f(C)$. Therefore, $Y \setminus f(C)$ is open in $Y$, and hence $f$ is a closed map.
Since $f$ is a continuous and closed map, it is a quotient map.
Suppose $X$ is a Hausdorff space and $q: X \rightarrow Y$ is a quotient map. Further suppose that $q$ is a closed map and that $q^{-1}(y)$ is compact for all $y \in Y$. Prove that $Y$ is Hausdorff.
Let $y_1, y_2$ be distinct points in $Y$. Since $q$ is onto, there exist $x_1, x_2 \in X$ such that $q(x_1) = y_1$ and $q(x_2) = y_2$. Suppose for contradiction that $Y$ is not Hausdorff, so that there exist no neighborhoods of $y_1$ and $y_2$ that are disjoint.
Then, for any open sets $U_1$ and $U_2$ containing $y_1$ and $y_2$, respectively, we have $U_1 \cap U_2 \neq \varnothing$. Since $q$ is a quotient map, $U_1 = q(V_1)$ and $U_2 = q(V_2)$ for some open sets $V_1$ and $V_2$ in $X$. Since $q(x_1) = y_1 \in U_1$, we have $x_1 \in V_1$, and similarly $x_2 \in V_2$. Therefore, $V_1 \cap V_2 \neq \varnothing$.
Since $X$ is Hausdorff, there exist disjoint open sets $W_1$ and $W_2$ in $X$ containing $x_1$ and $x_2$, respectively. Then, $q(W_1)$ and $q(W_2)$ are open sets in $Y$ containing $y_1$ and $y_2$, respectively. However, $q(W_1) \cap q(W_2) = q(W_1 \cap W_2) = \varnothing$, which contradicts our assumption that no neighborhoods of $y_1$ and $y_2$ are disjoint.
Therefore, $Y$ must be Hausdorff.
There was a theorem in $\S 7$ whose proof involved an infinite number of arbitrary choices. Which one was it? Rewrite the proof making explicit use of the choice axiom.
The theorem in $\S 7$ that involves an infinite number of arbitrary choices is the Tychonoff’s theorem, which states that the product of any collection of compact spaces is compact in the product topology.
To prove Tychonoff’s theorem using the choice axiom, we first note that the statement is equivalent to the assertion that the product of any collection of closed intervals $[a_i, b_i]$ is compact. This is because any compact space is homeomorphic to a closed and bounded interval in $\mathbb{R}$, and the product of closed intervals is homeomorphic to the product of the corresponding closed and bounded intervals.
We prove the theorem by contradiction. Assume that the product $\prod_{i \in I} [a_i, b_i]$ is not compact, so there exists an open cover ${U_\alpha}{\alpha \in A}$ with no finite subcover. Then, for each $i \in I$, let $F_i$ be the set of all $\alpha \in A$ such that the projection of $U\alpha$ onto the $i$-th factor of the product contains the point $a_i$. Since $[a_i, b_i]$ is compact, the projection of the product onto the $i$-th factor is a closed and bounded interval, and therefore, the set $F_i$ is finite.
Now, we use the axiom of choice to choose an element $\alpha_i \in A$ for each $i \in I$ such that $\alpha_i \notin F_i$. This is possible because $F_i$ is finite, and the complement of a finite set is infinite. Then, let $U = \prod_{i \in I} U_{\alpha_i}$ be the product of the open sets $U_{\alpha_i}$. Since $\alpha_i \notin F_i$ for each $i$, the projection of $U$ onto the $i$-th factor does not contain the point $a_i$. Hence, $U$ does not intersect the set $\prod_{i \in I} {a_i}$.
However, $\prod_{i \in I} [a_i, b_i]$ is compact, so the set $\prod_{i \in I} {a_i}$ is a closed subset of the product. Since $U$ is an open cover of the product with no finite subcover, it must intersect $\prod_{i \in I} {a_i}$, which is a contradiction. Therefore, the product $\prod_{i \in I} [a_i, b_i]$ is compact, and hence, Tychonoff’s theorem holds.
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