# crimboi[15] 序数拓扑(?)

##### 题目背景

本题竟然与xgn目前正在学习的序数有关, 真是可喜可贺!(crim boi在thu里有包含序数的课程吗?)

##### 题面

*If you don't remember what a topology is, please refer to this passage.*

Preliminary: Open sets in \mathbb R are just union (can be infinite) of some open intervals.

For a continuous function f:X\rightarrow \mathbb R, i.e. f^{-1}(V) is open in X for every open set V\subset\mathbb R, we define its support as follows:

\text{supp}f=\text{Cl}(\lbrace x:f(x)\ne 0\rbrace)

It's easy to demonstrate that on \mathbb R a compact set K is the support of some continuous function iff there's an open V s.t. \text{Cl}(V)=K.

Question: can you provide a counterexample of the statement above for general topological spaces?

As it's way too hard to directly answer it, we pursue the following sub-tasks, in which you may learn many practical tricks.

Topology on ordinal is defined by open sets being the union of some (\alpha,\beta), \lbrace x:x<\alpha\rbrace or \lbrace x:x>\alpha\rbrace. For example, in \omega+1, \lbrace x:x<1\rbrace=\lbrace0\rbrace is an open set, but \lbrace \omega\rbrace is not an open set. Ordinal spaces serves as an abundant source of counterexamples.

(i) A topological space X is called *first countable* iff given any x\in X, there is a countable family of open neighborhoods of x, namely U_1,U_2,... s.t. for any open neighborhood U of x, \exists U_i\subset U. Now prove \mathbb R is first countable, and given

\omega_1:=\min\lbrace x:|x|>|\omega|\rbrace

\omega_1+1 is not first countable (consider neighborhood of \omega_1).

(ii) In \omega_1+1, [0,\omega_1) is an open set and its closure is \omega_1+1. Using (i), prove \omega_1+1, which is the closure of an open set, is not a support of any continuous function from \omega_1+1 to \mathbb R.

(iii) Using well-ordering of an ordinal (or transfinite induction if you really want), prove \omega_1+1 is compact.

Proving three statements above, we can see that the first condition we provide fails. Yet, it's also not hard to strengthen the condition: on normal spaces, a compact set K is the support of some continuous function iff there's an open V and a countable family of closed sets C_i s.t. V=\bigcup C_i and \text{Cl}(V)=K.

However, as \omega_1+1 isn't first countable at \omega_1, it's impossible to find the countable closed family whose union is the open set we want. Moreover, as compact Hausdorff space is normal, we actually need to find a non-hausdorff space and say good-bye to ordinal space 🙁

先想再看提示哦

##### 提示

if ur big brain enough, just search "~First Countable + Countable + Compact + ~k_1-Hausdorff" on \pi-base, then the first result - One Point Compactification of the Rationals - would be the answer 😮

愿你无需看标答

##### 标答

感觉和构造搞笑二元函数的题差不多难, 不写标答了.

版权声明：

作者：HDD

链接：__https://blog.hellholestudios.top/archives/1202__

来源：Hell Hole Studios Blog

文章版权归作者所有，未经允许请勿转载。

## 共有 0 条评论