Complex2-Liu | 老爹 yinyang

One cannot turn back the hands of time.

如果努力的话, 也许, 生命比想象中要顽强.

点集拓扑复习 -- Locally Compact Hausdorff

在这篇文章里, 我们会复习一些点集拓扑的概念, 包括但不限于:

  1. 什么是正则 (regular) 和正规 (normal) 空间.
  2. 正则空间和正规空间的等价定义.
  3. 局部紧致空间的局部具有一些歧义, 但是在 Hausdorff 的时候就不会有这样的问题.
  4. 流形 is paracompact (仿紧).
$$ \newcommand{\ol}{\overline} \DeclareMathOperator{\id}{id} \DeclareMathOperator{\Hom}{Hom} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\norm}{\lVert}{\rVert} $$

正则空间和正规空间

设 $X$ 是一个拓扑空间, 我们称 $X$

  1. 正则, 如果我们能用开集来区分一个点和闭集.
  2. 正规, 如果我们能用开集来区分不相交的两个闭集.
Proposition (正则, 正规空间的等价定义)
  1. $X$ is regular if and only if for all $x \in U$ with $U$ open, there exists $V$ open such that $x \in V \subseteq \ol{V} \subseteq U$.

  2. $X$ is normal if and only if for all $A \subseteq E \subset X$ with $A$ closed and $E$ open, there exists $V$ open such that $A \subseteq V \subseteq \ol{V} \subseteq E$.

Proof. (2) Assume $X$ is normal. Choose $C, D$ open such that $A \subseteq C, E^c \subseteq D$ and $C \cap D = \varnothing$. Then $\ol{C} \cap D = \varnothing$, so $A \subseteq C \subseteq \ol{C} \subseteq E$.
Conversely, let $A, B$ be disjoint closed set, we can choose $V$ open such that $A \subseteq V \subseteq \ol{V} \subseteq B^c$. Then $V$ and $X - \ol{V}$ separate $A$ and $B$.
(1) 证明类似, 省略. $\square$

证明中用到了下述关键性质:

Lemma
If $A, B$ are open sets such that $A \cap B$ is empty, then $\ol{A} \cap B$ is again empty.

真是惭愧, 仔细回想, 这两个结论老夫当初学拓扑的时候肯定学过, 可以说是 regular 和 normal space 的等价定义, 然而不是今天刚好 review, 我真是几乎忘掉了.

局部紧致空间的歧义

$X$ is said to be locally (path) connected if every point admits a neighbourhood basis consisting entirely of open, (path) connected sets. 但是非常无语的是, $X$ is said to be locally compact if every point $x$ of $X$ has a compact neighbourhood. 有没有人可以告诉我, 为什么 locally compact 的定义要求存在一个紧致邻域就够了, 但是 locally connected 就要求存在一组邻域基呢?

不过幸运的是, 当 $X$ 是 Hausdorff 的时候, 存在一个紧致邻域和存在紧致邻域基是一回事, 因此不会产生歧义.

Proposition (Locally compact in Hausdorff space)

Assume $X$ is Hausdorff, the following are equivalent:

  1. every point of $X$ has a compact neighbourhood.
  2. every point of $X$ has a local base of compact neighbourhoods.
Proof. 回忆下述性质:
Assume (1) holds, let $x \in U \subseteq K \subseteq X$ with $K$ compact and $U$ open. One can wlog assume $X$ is compact (otherwise replace $X$ with $K$) and thus regular. Now $U$ is an open neighbourhood of $x$, we must find a compact neighbourhood of $x$ contained in $U$. Since $X$ is regular, the result then follows by Proposition (076b70b). $\square$
Proposition

Let $X$ be a locally compact Hausdorff space. For any basis $\CB$, the family

$$ \CB_c := \{U \in \CB: \ol{U} \text{ is compact}\} $$

is also a basis. 换句话说, 对于局部紧致的 Hausdorff 空间, 我们总是可以不妨假设它的拓扑基都有 compact closure.

Proof. We need to show that for all $x \in V$ with $V$ open, there is a $U \in \CB_c$ such that $x \in U \subseteq V$. By Proposition (9c11705), there is a compact neighbourhood $K_x$ of $x$ contained in $V$. Since $\mathcal{B}$ is a basis, there exists $U \in \mathcal{B}$ such that $x \in U \subseteq \mathrm{Int}(K_x)$. Note that $K_x$ is closed, so $\overline{U} \subseteq K_x$. Now $K_x$ is compact Hausdorff and $\overline{U}$ is its closed subset, so $\overline{U}$ is compact, i.e. $U \in \mathcal{B}_c$. $\square$
Corollary
Let $X$ be a topological space which is locally compact, Hausdorff and second countable. Then one can wlog choose its countable basis consisting of open sets with compact closures.

最后我们再给出一个定理, 该定理说明了任意流形都是仿紧的, 我们会在单位分解定理的证明中使用这个结论.

Theorem (manifold is paracompact)
Let $X$ as above. Then $X$ is paracompact (仿紧). In fact, each open cover has a countable, locally finite refinement consisting of open sets with compact closure.
Proof. See [2, Lemma 1.9] .
因为当 $i - j \ge 3$ 的时候, $$ G_{i+1} - \overline{G_{i-2}} \cap G_{j+1} - \overline{G_{j-2}} = \varnothing. $$ 所以最后构造出的 refinement 是 locally finite $\square$

References

1
Ronald Brown. Topology and Groupoids. http://groupoids.org.uk/pdffiles/topgrpds-e.pdf
2
Frank W. Warner. Foundations of Differentiable Manifolds and Lie Groups. GTM 94.