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# 数学代写|拓扑学代写TOPOLOGY代考|Noetherian Spaces

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## 数学代写|拓扑学代写TOPOLOGY代考|Noetherian Spaces

The upcoming result is very much used in algebraic geometry and commutative algebra.
Proposition 8.21 On an ordered set $(X, \leq)$ the following are equivalent:

1. every non-empty subset of $X$ contains maximal elements;
2. every countable ascending chain $\left{x_1 \leq x_2 \leq \cdots\right} \subset X$ stabilises (i.e. there’s an index $m \in \mathbb{N}$ such that $x_n=x_m$ for every $n \geq m$ ).

Proof For (1) $\Rightarrow(2)$ it’s enough to notice that any countable ascending chain $\left{x_1 \leq\right.$ $\left.x_2 \leq \cdots\right}$ contains a maximal element, say $x_m$, and therefore $x_n=x_m$ for every $n \geq m$

Let us prove (2) $\Rightarrow$ (1). By contradiction, suppose we have a non-empty $S \subset X$ with no maximal elements, i.e. ${y \in S \mid y>x} \neq \emptyset$ for every $x \in S$. By the axiom of choice there’s a function $f: S \rightarrow S$ such that $f(x)>x$ for every $x \in S$. Take $x_0 \in S$ : the ascending chain $\left{x_n=f^n\left(x_0\right) \mid n \in \mathbb{N}\right}$ does not stabilise.

Definition 8.22 A space is called Noetherian if every non-empty family of open sets has a maximal element for the inclusion.

By Proposition 8.21 a space is Noetherian if and only if every countable ascending chain stabilises.

Example 8.23 Let $\mathbb{K}$ be any field. The affine space $\mathbb{K}^n$ equipped with the Zariski topology (Example 3.11) is Noetherian. The proof is a simple consequence of Hilbert’s basis theorem, ${ }^1$ and as such we leave it to lecture courses on algebraic geometry and commutative algebra.

## 数学代写|拓扑学代写TOPOLOGY代考|A Long Exercise: Tietze’s Extension Theorem

Recall that a space is called normal when it is Hausdorff and disjoint closed sets have disjoint neighbourhoods. The exercises at the end of the section, solved in the given order, provide a proof of the following two results.

Lemma 8.29 (Urysohn’s lemma) Let $A, C$ be disjoint closed sets in a normal space $X$. There exists a continuous map $f: X \rightarrow[0,1]$ such that $f(x)=0$ when $x \in A$ and $f(x)=1$ when $x \in C$.

Theorem 8.30 (Tietze extension) Let $B$ be closed in a normal space $X, J \subset \mathbb{R} a$ convex subspace and $f: B \rightarrow J$ a continuous map. Then there exists a continuous map $g: X \rightarrow J$ such that $g(x)=f(x)$ for every $x \in B$.

For metric spaces Urysohn’s lemma is easy to prove: it suffices to recycle the argument of Proposition 7.31.

Let us remark that Lemma 8.29 is a special case of Theorem 8.30 when $J=[0,1]$ and $B=A \cup C$. The classical proof of Theorem 8.30 , for which we suggest consulting [Mu00, Du66], uses Urysohn’s lemma and the completeness of the space $B C(X, \mathbb{R})$ of continuous and bounded real functions on $X$.

## 数学代写|拓扑学代写TOPOLOGY代考|Noetherian Spaces

$X$的每个非空子集都包含最大元素;

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