# 测试逻辑[番外] 超限归纳在拓扑上的一个应用

此文作讲解的不是数理逻辑, 而是测试逻辑中一个重要定理---超限归纳在拓扑的一个基本定理上的应用(其实还顺带介绍了不少拓扑). 当然, 在本篇的之后可能会讲到佐恩引理, 那么这些用超限归纳解题的过程都能被简化, 这也是大多数教科书所用的做法. 不过归根结底本文目的是让读者见识到数理逻辑并非是无用的"玩具"; 相反, 超限归纳在较为初等的学科, 如拓扑学上也能大显身手.

*另外由于本文是番外, 笔者将使用他比较熟悉的英文以及排版较为简单的markdown进行写作.*

##### 1 Preliminary

**1.1** A topological space is a set S equipped with a \tau\subset P(S) such that

\emptyset\in\tau, S\in\tau;

\forall x,y\in\tau(x\cap y\in\tau);

\forall A\subset\tau(\cup A\in\tau).

The elements of \tau are called open sets. On the contrary, we call X\subset S a closed set iff S-X\in\tau. Note that it's easy to prove the intersection of finitely many open sets is still an open set; the union of finitely many closed sets is a closed set, and the intersection of closed sets is a closed set.

**1.2** A normal space is a topological space S such that for any two disjoint closed sets A,B, there two disjoint open sets U,V such that A\subset U, B\subset V.

**1.3** For an X\subset S, we define the closure of X as

\text{Cl}(X)=\cap\lbrace A:A\text{ is closed and }X\subset A\rbrace

**1.4** A set of A\subset\tau is called an open covering of S iff \cup A=S. We often denote A as \lbrace U_i\rbrace_{i\in I}.

**1.5** A locally finite covering of S is a covering \lbrace U_i\rbrace_{i\in I} such that for every x\in S, there is an open set X such that X only intersects with finite U_i. In other words, there is a finite J\subset I such that

X\cap\left(\mathop{\cup}\limits_{i\in I-J}U_i\right)=\emptyset

##### 2 An important result in normal spaces

**2.1** For any non-empty open set U in a normal space S and a closed set A\subset U, we can find an open set V such that A\subset V\subset\text{Cl}(V)\subset U.

**Proof** Since U is open, S-U is closed. As A\subset U, A and S-U are disjoint. By normality, there are disjoint open V,W such that A\subset V and S-U \subset W. As V, W are disjoint, V\subset S-W\subset U. Now since W is open, S-W is closed, and V\subset S-W, so \text{Cl}(V)\subset S-W\subset U.

##### 3 An application of transfinite induction---Shrinking Lemma

Now we present a lemma crucial to prove the magical existence of partition of unity---shrinking lemma.

**3.1**(Shrinking Lemma) For a locally-finite covering \lbrace U_i\rbrace_{i\in I} of a normal space S, there's a covering \lbrace V_i\rbrace_{i\in I} such that \forall i\in I(V_i\subset\text{Cl}(V_i)\sub U_i).

Before diving into this lemma's proof, we firstly investigate the finite situation.

Let's start from the most simple situation: S is covered by two open sets: U_1,U_2. Then we can extend U_1-U_2, a closed subset of U_1, to some open V_1 such that \text{Cl}(V_1)\sub U_1 and V_1\cup U_2=S.

For a finite open covering \lbrace U_1,...,U_n\rbrace, we take U_1 and \mathop{\cup}\limits_{i=2}^n U_i as a binary covering. In this way, we can find V_1\subset\text{Cl}(V_1)\subset U_1, and the finite covering becomes \lbrace V_1,U_2,...,U_n\rbrace. Then we take U_2 and V_1\cup\left(\mathop{\cup}\limits_{i=3}^n U_i\right) as a binary covering, and we can similarly find V_2. After finite steps, we can turn the initial covering into \lbrace V_1,...,V_n\rbrace satisfying the conditions.

\lbrace V_1,...,V_n\rbrace is a covering because we check this after every step turing U_i into V_i. However, when I is infinite, this kind of "checking" won't work. Yet we can still inherit the idea of induction and utilize a powerful tool against infinity---transfinite induction.

**Proof of 3.1** Admitting axiom of choice, we take a well ordering of I. Suppose \text{Ord}(I)=\gamma, we prove that there is a sequence (V_0,V_1,...) such that for all \alpha\le\gamma

(\cup\lbrace V_\beta:\beta<\alpha\rbrace)\cup(\cup\lbrace U_\beta:\beta\ge\alpha\rbrace)=S

by induction on \alpha.

Firstly when \alpha=0, the proposition is immediate as \cup\lbrace U_\beta:\beta\ge0\rbrace is just the initial covering.

For a successor ordinal \alpha+1, we prove that the proposition holds if it holds for \alpha. To be particular, when

(\cup\lbrace V_\beta:\beta<\alpha\rbrace)\cup(\cup\lbrace U_\beta:\beta\ge\alpha\rbrace)=S

we want to prove

(\cup\lbrace V_\beta:\beta<\alpha\rbrace)\cup V_{\alpha}\cup(\cup\lbrace U_\beta:\beta\ge\alpha+1\rbrace)=S

To prove this, we just imitate what we did in finite situations: take U_\alpha and (\cup\lbrace V_\beta:\beta<\alpha\rbrace)\cup(\cup\lbrace U_\beta:\beta\ge\alpha+1\rbrace) as a binary covering, then there's a V_{\alpha}\sub\text{Cl}(V_\alpha)\sub U_\alpha such that

(\cup\lbrace V_\beta:\beta<\alpha\rbrace)\cup(\cup\lbrace U_\beta:\beta\ge\alpha+1\rbrace)\cup V_{\alpha}=S

For a non-zero limit ordinal \alpha, we prove that when for all \delta<\alpha

(\cup\lbrace V_\beta:\beta<\delta\rbrace)\cup(\cup\lbrace U_\beta:\beta\ge\delta\rbrace)=S

the following holds:

(\cup\lbrace V_\beta:\beta<\alpha\rbrace)\cup(\cup\lbrace U_\beta:\beta\ge\alpha\rbrace)=S

If not, take an x\in S-(\cup\lbrace V_\beta:\beta<\alpha\rbrace)\cup(\cup\lbrace U_\beta:\beta\ge\alpha\rbrace), then we have \forall\beta<\alpha(x\notin V_\beta) and \forall\beta\ge\alpha(x\notin U_\beta).

By locally finiteness, take an open O_x\ni x such that O_x intersects with finitely many U_i. Now if x is contained in some U_i, U_i must intersects with O_x, so x is only contained in finitely many U_i, namely \lbrace U_{\delta_1},...,U_{\delta_n}\rbrace. By the above argument, \forall 1\le i\le n(\delta_i<\alpha). Suppose \delta=\max\limits_{1\le i\le n}\lbrace\delta_i\rbrace, as \alpha is a non-zero limit ordinal, \delta+1<\alpha, and the following assumed proposition leads to a contradiction:

(\cup\lbrace V_\beta:\beta<\delta+1\rbrace)\cup(\cup\lbrace U_\beta:\beta\ge\delta+1\rbrace)=S

Since \forall\beta<\alpha(x\notin V_\beta), x\notin\cup\lbrace V_\beta:\beta<\delta+1\rbrace, so x\in \cup\lbrace U_\beta:\beta\ge\delta+1\rbrace. Yet \delta is the largest ordinal such that U_\delta\ni x, which is contradictory.

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作者：HDD

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来源：Hell Hole Studios Blog

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