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问题 1.

Problem 2. [A criterion for paracompactness.]
(a). Let $M$ be any topological space which is locally Euclidean. Prove that $M$ is second countable iff $M$ has a countable atlas.
[Pedantically, in Lecture #1 we only defined the notation of an “atlas” when $M$ is connected and Hausdorff; however the same defnition applies to any locally Euclidean topological space.]
(b). Let $M$ be a connected Hausdorff space which is locally Euclidean. Prove that $M$ is paracompact iff $M$ has a countable atlas.


(a) Let’s prove the forward direction first: if $M$ is second countable, then it has a countable atlas.

Assume that $M$ is second countable, i.e., there exists a countable base $\mathcal{B}$ for the topology on $M$. We want to construct a countable atlas for $M$.

For each $x \in M$, since $M$ is locally Euclidean, there exists a neighborhood $U_x$ of $x$ that is homeomorphic to an open subset of Euclidean space $\mathbb{R}^n$. Let $\mathcal{B}_x$ be a countable base for the topology on $U_x$. Since $\mathbb{R}^n$ is second countable (it has a countable base), we can choose a countable subset $\mathcal{B}_x’ \subseteq \mathcal{B}_x$ that is a base for $U_x$. Note that $\mathcal{B}_x’$ is also a base for the subspace topology on $U_x$ induced by $M$.

Now, consider the collection of all pairs $(U_x, \varphi_x)$, where $U_x$ is a neighborhood of $x$ in $M$ and $\varphi_x: U_x \to \mathbb{R}^n$ is a homeomorphism. Define the set
$$\mathcal{A} = {(U_x, \varphi_x) : x \in M}.$$

We claim that $\mathcal{A}$ is a countable atlas for $M$. To show this, we need to verify the following properties:

$\mathcal{A}$ covers $M$: For any $x \in M$, since $M$ is locally Euclidean, there exists $(U_x, \varphi_x) \in \mathcal{A}$ such that $x \in U_x$. Therefore, $\mathcal{A}$ covers $M$.

Compatibility: Let $(U_x, \varphi_x)$ and $(U_y, \varphi_y)$ be two charts in $\mathcal{A}$ such that $U_x \cap U_y \neq \varnothing$. We need to show that the transition function
$$\varphi_y \circ \varphi_x^{-1} : \varphi_x(U_x \cap U_y) \to \varphi_y(U_x \cap U_y)$$
is a homeomorphism. Since $M$ is locally Euclidean, $U_x$ and $U_y$ are both homeomorphic to open subsets of $\mathbb{R}^n$, and the transition function can be identified with the restriction of a homeomorphism between open subsets of $\mathbb{R}^n$. Therefore, the transition function is a homeomorphism.

Countability: We need to show that $\mathcal{A}$ is countable. Since $\mathcal{B}$ is a countable base for the topology on $M$, the set
$$\mathcal{B}’ = {(U_x, \varphi_x) \in \mathcal{A} : U_x \in \mathcal{B}}$$
is a subset of $\mathcal{A}$ and is countable. Moreover, every chart in $\mathcal{A}$ is uniquely determined by its associated open set $U_x$, so $\mathcal{B}’$ is in one-to-one correspondence with $\mathcal{B}$. Hence, $\mathcal{A}$ is countable.

Therefore, $M$ has a countable atlas.

Now

问题 2.

Problem 3. [Invariance of dimension.] Brouwer’s Theorem on invariance of dimension states: If nonempty open sets $U \subset \mathbb{R}^{d_1}$ and $V \subset \mathbb{R}^{d_1}$ are homeomorphic, then $d_1=d_2$. (Cf., e.g., Hatcher, [7, Thm. 2.26].) Using this result, prove the following: If $M$ is a connected Hausdorff space for which every point has an open neighborhood $U$ which is homeomorphic to an open subset $\Omega$ of $\mathbb{R}^d$ for some $d \in \mathbb{Z}_{\geq 1}$ (which apriori may depend on $U$ ), then in fact all the dimensions $d$ appearing must be one and the same.
(Thus, in Def. 1 in Lecture #1, we would not obtain any new objects if we modify the definition so that the dimension is allowed to depend on $U$. .)

To prove that all dimensions appearing in the given conditions are the same, we can use Brouwer’s Theorem on the invariance of dimension. Let’s proceed with the proof.

Assume that $M$ is a connected Hausdorff space such that every point has an open neighborhood $U$ homeomorphic to an open subset $\Omega \subset \mathbb{R}^{d}$ for some $d \in \mathbb{Z}_{\geq 1}$. We want to show that all the dimensions $d$ appearing in this condition are the same.

Suppose, for the sake of contradiction, that there exist $d_1$ and $d_2$ such that $d_1 < d_2$. Let $p \in M$ be any point, and let $U_p$ be an open neighborhood of $p$ that is homeomorphic to an open subset $\Omega \subset \mathbb{R}^{d_1}$. Since $M$ is connected, there exists a path-connected open neighborhood $V_p \subset U_p$ of $p$.

Now, consider the collection of all such path-connected open neighborhoods $V_p$ for all $p \in M$. This collection forms an open cover for $M$. Since $M$ is a connected space, this cover has a finite subcover. Let $V_{p_1}, V_{p_2}, \ldots, V_{p_n}$ be a finite subcover.

Consider the intersection $U_{p_1} \cap U_{p_2} \cap \ldots \cap U_{p_n}$. This intersection is an open subset of $\mathbb{R}^{d_1}$. Let $U$ be the union of all $U_{p_i}$’s, which is an open neighborhood of $M$. Similarly, let $\Omega$ be the union of all $\Omega_{p_i}$’s, which is an open subset of $\mathbb{R}^{d_1}$. Since $\mathbb{R}^{d_1}$ and $\mathbb{R}^{d_2}$ are not homeomorphic when $d_1 < d_2$ by Brouwer’s Theorem, we have a contradiction since $U$ and $\Omega$ are assumed to be homeomorphic.

Hence, we can conclude that if $M$ is a connected Hausdorff space where every point has an open neighborhood homeomorphic to an open subset of $\mathbb{R}^{d}$ for some $d \in \mathbb{Z}_{\geq 1}$, then all the dimensions $d$ appearing must be the same.

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