p******n
发帖数: 46 | 1 本人拓扑太弱了，没有思路，请高手指点一下：
Under what conditions does a metrizable space have a metrizable
compactification?
谢谢了！ |
H****h
发帖数: 1037 | 2 不知道这个条件够不够？对于任何小的正数a，都存在有限个半径为a的球覆盖。
【在 p******n 的大作中提到】
: 本人拓扑太弱了，没有思路，请高手指点一下：
: Under what conditions does a metrizable space have a metrizable
: compactification?
: 谢谢了！
|
r********e
发帖数: 103 | 3 separable (or second countable)? |
p******n
发帖数: 46 | 4 Why does it need to be separable?
metrizable space => normal space => completely regular space => can be
embedded into [0,1]^J which is compact => has a compactification
where should we use the property "separable"?
Another direction, a separable metrizable space is compact.
Please help me figure it out. |
r********e
发帖数: 103 | 5 Note that all second-countable spaces are separable. A metric space is
separable if and only if it is second-countable. The two conditions are
equivalent here.
1) A compact metrizable space is second countable. Its subspaces are second
countable too.
2) A second countable metrisable space is homeomorphic
to a subspace of [0,1]^J, which leads to a metrizable compactification. |
p******n
发帖数: 46 | 6 2)中的 2nd countable必要吗？为什么？
metrizable space是normal space.因此必然是 completely regular space. 一个
completely regular space is homeomorphic to a subspace of [0,1]^J.
为什么还要 2nd countable呢？我的推理有问题吗？
second
【在 r********e 的大作中提到】
: Note that all second-countable spaces are separable. A metric space is
: separable if and only if it is second-countable. The two conditions are
: equivalent here.
: 1) A compact metrizable space is second countable. Its subspaces are second
: countable too.
: 2) A second countable metrisable space is homeomorphic
: to a subspace of [0,1]^J, which leads to a metrizable compactification.
|
r********e
发帖数: 103 | 7 metrizable是completely regular Hausdorff space 或者说是Tychonoff space没错，
但是这个只保证homeomorphic to a subspace of the cube [0,1]^I, rather than
the Hilbert cube [0,1]^N. Notice that the cube may not be metrizable since
the index set I may not be countable (think about the cantor cube).
Actually from Urysohn Metrization Theorem, we have the following:
For a Hausdorff space X, TFAE:
1. X can be embedded in the Hilbert cube
2. X is separable metrizable space
3. X is regular and second countable
You can see |
p******n
发帖数: 46 | |
