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定义 1.1. 设$X$为$T_2$空间, 若任意开覆盖有局部有限加细, 则称$X$为仿紧空间.
命题 1.2. 流形是仿紧空间.
证: 取流形$M^n$的坐标图册$\{(U_\alpha,\varphi_\alpha)\},$
首先可将开覆盖加细使得每个开集落在某个坐标邻域$U_\alpha$内(取交即可),
再由流形的$C_2$性, 开覆盖有可数子覆盖$\{V_i\}.$ 设每个$V_i\subset U_i,$
$(U_i,\varphi_i)$为坐标邻域且允许对某个$i\neq j$有$U_i=U_j.$
将每个$V_i$拉回到$\mathbb{R}^n$上,
那么开集$\varphi_i(V_i)$有穷竭序列$K_{i,1}\subset K_{i,2}\subset \cdots,$
每个$K_{i,j}$为紧集. 接下来构造我们想要的局部有限加细:
$$W_i:=V_i\setminus \bigcup_{j=1}^{i-1} \varphi_{j}^{-1}(K_{j,i})$$
那么$\{W_i\}$满足要求. 首先每个$W_i$确实是开集,
因为$\varphi_i$为同胚而$K_{j,i}$为紧集, 紧集的连续像仍是紧集,
同时在$T_2$空间是闭的. 它当然是覆盖$\{V_i\}$的加细,
也是初始开覆盖的加细. 最后我们说明它确实是局部有限的开覆盖.
$\,\forall\,x\in M,$ $\,\exists\,V_i \ni x.$
令$V_a$为其中指标最小的开集,
那么由于$\varphi_{j}^{-1}(K_{j,i})\subset V_{j},$
$\,\forall\,j< a,$ $x\notin V_j,$ 从而$x\in W_a.$
然而又由$V_a$有穷竭序列, 存在$x$的小邻域$O_x,$ $\,\exists\,b>a,$
$\,\forall\,k> b,$ $O_x\subset \varphi_{a}^{-1}(K_{a,k}),$
因此$O_x\cap W_k=\varnothing.$ 从而$O_x$至多与$b$个开集有交,
于是$\{W_i\}$是开覆盖的局部有限加细.
参考: https://mathoverflow.net/a/96783
文章最后更新于 2021-10-28 19:43:17
Definition 1.1. Let $X$ be $T_2$ space. If any open cover has local finite thinning, then $X$ is called a paracompact space.
Proposition 1.2. Manifolds are paracompact spaces.
Certificate: Get the coordinate atlas $\{(U_\alpha,\varphi_\alpha)\},$ of the manifold $M^n$
First, the open coverage can be refined so that each open cluster falls within a certain coordinate neighborhood $U_\alpha$ (just take the intersection),
According to the $C_2$ property of the manifold, the open cover has countable sub-covers $\{V_i\}.$. Suppose each $V_i\subset U_i,$
$(U_i,\varphi_i)$ is a coordinate neighborhood and allows $U_i=U_j.$ for a certain $i\neq j$
Pull each $V_i$ back onto $\mathbb{R}^n$,
Then the open set $\varphi_i(V_i)$ has an exhaustive sequence $K_{i,1}\subset K_{i,2}\subset \cdots,$
Each $K_{i,j}$ is a compact set. Next, we construct the local finite refinement we want:
$$W_i:=V_i\setminus \bigcup_{j=1}^{i-1} \varphi_{j}^{-1}(K_{j,i})$$
Then $\{W_i\}$ meets the requirements. First, each $W_i$ is indeed an open set,
Because $\varphi_i$ is a homeomorphism and $K_{j,i}$ is a compact set, the continuous image of a compact set is still a compact set,
At the same time, it is closed in $T_2$ space. It certainly covers the thinning of $\{V_i\}$,
It is also a refinement of the initial open coverage. Finally, we show that it is indeed a locally limited open coverage.
$\,\forall\,x\in M,$ $\,\exists\,V_i \ni x.$
Let $V_a$ be the open set with the smallest index,
Then because $\varphi_{j}^{-1}(K_{j,i})\subset V_{j},$
$\,\forall\,j< a,$ $x\notin V_j,$ thus $x\in W_a.$
However, since $V_a$ has an exhaustive sequence, there is a small neighborhood $O_x,$ $\,\exists\,b>a,$ of $x$.
$\,\forall\,k> b,$ $O_x\subset \varphi_{a}^{-1}(K_{a,k}),$
Therefore $O_x\cap W_k=\varnothing.$ and thus $O_x$ intersect at most $b$ open sets,
So $\{W_i\}$ is a local finite thinning with open coverage.
Reference: https://mathoverflow.net/a/96783
The article was last updated on 2021-10-28 19:43:17
