逆数学导论（5）

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Formalizing Hahn's Embedding Theorem

哈恩嵌入定理的形式化

To state Hahn's embedding theorem in second order arithmetic, we need to define Hahn subgroups and isomorphisms between ordered abelian groups and Hahn subgroups.

为在二阶算术中表述哈恩嵌入定理，需先定义哈恩子群 及有序交换群与哈恩子群间的同构。

Definitions

定义

Let ( T , ≤ T ) (T, \leq_T) (T,≤T) be a linear order, and let { K t } t ∈ T \{K_t\}{t\in T} {Kt}t∈T be a sequence of Archimedean ordered abelian groups indexed by T T T. A **subgroup of ∑ t ∈ T K t \sum{t\in T} K_t ∑t∈TKt indexed by I I I** is a sequence of functions F = { f i : T → ⋃ t ∈ T K t } i ∈ I F = \{f_i: T \to \bigcup_{t\in T} K_t\}_{i\in I} F={fi:T→⋃t∈TKt}i∈I such that:

设 ( T , ≤ T ) (T, \leq_T) (T,≤T) 为全序， { K t } t ∈ T \{K_t\}{t\in T} {Kt}t∈T 为以 T T T 为指标集的阿基米德有序交换群序列。**以 I I I 为指标的 ∑ t ∈ T K t \sum{t\in T} K_t ∑t∈TKt 的子群** 是函数序列 F = { f i : T → ⋃ t ∈ T K t } i ∈ I F = \{f_i: T \to \bigcup_{t\in T} K_t\}_{i\in I} F={fi:T→⋃t∈TKt}i∈I，满足：

f i ( t ) ∈ K t f_i (t) \in K_t fi(t)∈Kt for all i ∈ I i \in I i∈I and t ∈ T t \in T t∈T;

对所有 i ∈ I i \in I i∈I 和 t ∈ T t \in T t∈T， f i ( t ) ∈ K t f_i (t) \in K_t fi(t)∈Kt；
There exists i 0 ∈ I i_0 \in I i0∈I such that f i 0 ( t ) = 0 K t f_{i_0}(t) = 0_{K_t} fi0(t)=0Kt for all t ∈ T t \in T t∈T (additive identity);

存在 i 0 ∈ I i_0 \in I i0∈I 使得对所有 t ∈ T t \in T t∈T， f i 0 ( t ) = 0 K t f_{i_0}(t) = 0_{K_t} fi0(t)=0Kt（加法单位元）；
For all i ∈ I i \in I i∈I, there exists j ∈ I j \in I j∈I such that f j ( t ) = − f i ( t ) f_j (t) = -f_i (t) fj(t)=−fi(t) for all t ∈ T t \in T t∈T (additive inverse);

对所有 i ∈ I i \in I i∈I，存在 j ∈ I j \in I j∈I 使得对所有 t ∈ T t \in T t∈T， f j ( t ) = − f i ( t ) f_j (t) = -f_i (t) fj(t)=−fi(t)（加法逆元）；
For all i , j ∈ I i, j \in I i,j∈I, there exists k ∈ I k \in I k∈I such that f k ( t ) = f i ( t ) + K t f j ( t ) f_k (t) = f_i (t) +_{K_t} f_j (t) fk(t)=fi(t)+Ktfj(t) for all t ∈ T t \in T t∈T (closure under addition);

对所有 i , j ∈ I i, j \in I i,j∈I，存在 k ∈ I k \in I k∈I 使得对所有 t ∈ T t \in T t∈T， f k ( t ) = f i ( t ) + K t f j ( t ) f_k (t) = f_i (t) +_{K_t} f_j (t) fk(t)=fi(t)+Ktfj(t)（对加法封闭）；
For all i ≠ j ∈ I i \neq j \in I i=j∈I, there exists t ∈ T t \in T t∈T such that f i ( t ) ≠ f j ( t ) f_i (t) \neq f_j (t) fi(t)=fj(t) (distinctness).

对所有 i ≠ j ∈ I i \neq j \in I i=j∈I，存在 t ∈ T t \in T t∈T 使得 f i ( t ) ≠ f j ( t ) f_i (t) \neq f_j (t) fi(t)=fj(t)（元素互异）。

F F F has the structure of an abelian group; to be anordered abelian group , it must satisfy an additional order condition: If the set { t ∣ f i ( t ) ≠ 0 K t } \{t \mid f_i (t) \neq 0_{K_t}\} {t∣fi(t)=0Kt} is well-ordered for all i ∈ I i \in I i∈I, define theorder ≤ F \leq_F ≤Fby f i < F f j f_i <F f_j fi<Ffj if and only if f i ( t 0 ) < K t 0 f j ( t 0 ) f_i (t_0) <{K_{t_0}} f_j (t_0) fi(t0)<Kt0fj(t0), where t 0 t_0 t0 is the T T T-minimal element of { t ∣ ( f i − f j ) ( t ) ≠ 0 K t } \{t \mid (f_i - f_j)(t) \neq 0_{K_t}\} {t∣(fi−fj)(t)=0Kt}. This order is valid if it respects addition (i.e., f i < F f j f_i <_F f_j fi<Ffj implies f i + F f k < F f j + F f k f_i +F f_k <F f_j +F f_k fi+Ffk<Ffj+Ffk for all k ∈ I k \in I k∈I).
F F F 具备交换群结构；若要成为有序交换群 ，还需满足序条件：若对所有 i ∈ I i \in I i∈I， { t ∣ f i ( t ) ≠ 0 K t } \{t \mid f_i (t) \neq 0{K_t}\} {t∣fi(t)=0Kt} 是良序集，则定义序 ≤ F \leq_F ≤F： f i < F f j f_i <F f_j fi<Ffj 当且仅当在 { t ∣ ( f i − f j ) ( t ) ≠ 0 K t } \{t \mid (f_i - f_j)(t) \neq 0{K_t}\} {t∣(fi−fj)(t)=0Kt} 的 T T T- 极小元 t 0 t_0 t0 处，有 f i ( t 0 ) < K t 0 f j ( t 0 ) f_i (t_0) <{K{t_0}} f_j (t_0) fi(t0)<Kt0fj(t0)。若该序对加法满足单调性（即 f i < F f j f_i <_F f_j fi<Ffj 可推出对所有 k ∈ I k \in I k∈I， f i + F f k < F f j + F f k f_i +_F f_k <_F f_j +_F f_k fi+Ffk<Ffj+Ffk），则 F F F 是有序交换群。

AHahn subgroup is a subgroup of ∑ t ∈ T K t \sum_{t\in T} K_t ∑t∈TKt that is an ordered abelian group (as above) and satisfies thecut property : For every t 0 ∈ T t_0 \in T t0∈T and i ∈ I i \in I i∈I, there exists j ∈ I j \in I j∈I such that f j ( t ) = f i ( t ) f_j (t) = f_i (t) fj(t)=fi(t) if t < T t 0 t <T t_0 t<Tt0, and f j ( t ) = 0 K t f_j (t) = 0{K_t} fj(t)=0Kt otherwise.
哈恩子群 是 ∑ t ∈ T K t \sum_{t\in T} K_t ∑t∈TKt 的子群，满足：（1）是上述有序交换群；（2）截断性质 ：对所有 t 0 ∈ T t_0 \in T t0∈T 和 i ∈ I i \in I i∈I，存在 j ∈ I j \in I j∈I 使得对所有 t ∈ T t \in T t∈T，若 t < T t 0 t <T t_0 t<Tt0，则 f j ( t ) = f i ( t ) f_j (t) = f_i (t) fj(t)=fi(t)；否则 f j ( t ) = 0 K t f_j (t) = 0{K_t} fj(t)=0Kt。

An ordered abelian group G G G is**isomorphic to a Hahn subgroup of ∑ t ∈ T K t \sum_{t\in T} K_t ∑t∈TKt**if there exists a Hahn subgroup F F F of ∑ t ∈ T K t \sum_{t\in T} K_t ∑t∈TKt indexed by G G G such that:

若存在以 G G G 为指标集的 ∑ t ∈ T K t \sum_{t\in T} K_t ∑t∈TKt 的哈恩子群 F F F，满足以下条件，则称有序交换群 G G G同构于 ∑ t ∈ T K t \sum_{t\in T} K_t ∑t∈TKt 的哈恩子群：

f 0 G ( t ) = 0 K t f_{0_G}(t) = 0_{K_t} f0G(t)=0Kt for all t ∈ T t \in T t∈T;

对所有 t ∈ T t \in T t∈T， f 0 G ( t ) = 0 K t f_{0_G}(t) = 0_{K_t} f0G(t)=0Kt；
f g + G h ( t ) = f g ( t ) + K t f h ( t ) f_{g +G h}(t) = f_g (t) +{K_t} f_h (t) fg+Gh(t)=fg(t)+Ktfh(t) for all g , h ∈ G g, h \in G g,h∈G and t ∈ T t \in T t∈T;

对所有 g , h ∈ G g, h \in G g,h∈G 和 t ∈ T t \in T t∈T， f g + G h ( t ) = f g ( t ) + K t f h ( t ) f_{g +G h}(t) = f_g (t) +{K_t} f_h (t) fg+Gh(t)=fg(t)+Ktfh(t)；
f − g ( t ) = − f g ( t ) f_{-g}(t) = -f_g (t) f−g(t)=−fg(t) for all g ∈ G g \in G g∈G and t ∈ T t \in T t∈T;

对所有 g ∈ G g \in G g∈G 和 t ∈ T t \in T t∈T， f − g ( t ) = − f g ( t ) f_{-g}(t) = -f_g (t) f−g(t)=−fg(t)；
g < G h ↔ f g < F f h g <_G h \leftrightarrow f_g <_F f_h g<Gh↔fg<Ffh for all g , h ∈ G g, h \in G g,h∈G.

对所有 g , h ∈ G g, h \in G g,h∈G， g < G h ↔ f g < F f h g <_G h \leftrightarrow f_g <_F f_h g<Gh↔fg<Ffh。

Main Theorem III

主定理 3
Hahn's Embedding Theorem (formalized): For every ordered abelian group G G G, there exists a linear order T T T and a sequence of Archimedean ordered abelian groups { K t } t ∈ T \{K_t\}{t\in T} {Kt}t∈T such that G G G is isomorphic to a Hahn subgroup of ∑ t ∈ T K t \sum{t\in T} K_t ∑t∈TKt.
哈恩嵌入定理 （形式化表述）：对每个有序交换群 G G G，存在全序 T T T 和阿基米德有序交换群序列 { K t } t ∈ T \{K_t\}{t\in T} {Kt}t∈T，使得 G G G 同构于 ∑ t ∈ T K t \sum{t\in T} K_t ∑t∈TKt 的哈恩子群。

Claim : Hahn's embedding theorem implies ACA 0 \text {ACA}_0 ACA0 over RCA 0 \text {RCA}_0 RCA0.
断言：在 RCA 0 \text {RCA}_0 RCA0 上，哈恩嵌入定理可推出 ACA 0 \text {ACA}_0 ACA0。

Proof Let f : N → N f: \mathbb {N} \to \mathbb {N} f:N→N be an arbitrary function. Define G G G as in the proof of Theorem III.2---an ordered abelian group constructed from f f f. By Hahn's embedding theorem, there exists an isomorphism { f g } g ∈ G \{f_g\}{g\in G} {fg}g∈G between G G G and a Hahn subgroup F F F of ∑ t ∈ T K t \sum{t\in T} K_t ∑t∈TKt.
证明设 f : N → N f: \mathbb {N} \to \mathbb {N} f:N→N 为任意函数，按定理 3.2 的构造定义有序交换群 G G G。由哈恩嵌入定理，存在同构 { f g } g ∈ G \{f_g\}{g\in G} {fg}g∈G 将 G G G 映射到 ∑ t ∈ T K t \sum{t\in T} K_t ∑t∈TKt 的某哈恩子群 F F F。

We claim that x 2 n ≈ x 2 n + 1 ↔ ∀ t ∈ T ( f x 2 n + 1 ( t ) ≠ 0 K t → f x 2 n ( t ) ≠ 0 K t ) x_{2n} \approx x_{2n+1} \leftrightarrow \forall t\in T\left (f_{x_{2n+1}}(t) \neq 0_{K_t} \to f_{x_{2n}}(t) \neq 0_{K_t}\right) x2n≈x2n+1↔∀t∈T(fx2n+1(t)=0Kt→fx2n(t)=0Kt):

断言： x 2 n ≈ x 2 n + 1 ↔ ∀ t ∈ T ( f x 2 n + 1 ( t ) ≠ 0 K t → f x 2 n ( t ) ≠ 0 K t ) x_{2n} \approx x_{2n+1} \leftrightarrow \forall t\in T\left (f_{x_{2n+1}}(t) \neq 0_{K_t} \to f_{x_{2n}}(t) \neq 0_{K_t}\right) x2n≈x2n+1↔∀t∈T(fx2n+1(t)=0Kt→fx2n(t)=0Kt)。

If x 2 n ≈ x 2 n + 1 x_{2n} \approx x_{2n+1} x2n≈x2n+1, then there exists an odd prime p m p_m pm such that p m x 2 n = x 2 n + 1 p_m x_{2n} = x_{2n+1} pmx2n=x2n+1 (since n n n is in the range of f f f). By the isomorphism property, p m f x 2 n ( t ) = f x 2 n + 1 ( t ) p_m f_{x_{2n}}(t) = f_{x_{2n+1}}(t) pmfx2n(t)=fx2n+1(t) for all t ∈ T t \in T t∈T. Thus f x 2 n + 1 ( t ) ≠ 0 K t f_{x_{2n+1}}(t) \neq 0_{K_t} fx2n+1(t)=0Kt implies f x 2 n ( t ) ≠ 0 K t f_{x_{2n}}(t) \neq 0_{K_t} fx2n(t)=0Kt;
若 x 2 n ≈ x 2 n + 1 x_{2n} \approx x_{2n+1} x2n≈x2n+1，则存在奇素数 p m p_m pm 使得 p m x 2 n = x 2 n + 1 p_m x_{2n} = x_{2n+1} pmx2n=x2n+1（因 n n n 在 f f f 的值域中）。由同构性质，对所有 t ∈ T t \in T t∈T， p m f x 2 n ( t ) = f x 2 n + 1 ( t ) p_m f_{x_{2n}}(t) = f_{x_{2n+1}}(t) pmfx2n(t)=fx2n+1(t)，故 f x 2 n + 1 ( t ) ≠ 0 K t f_{x_{2n+1}}(t) \neq 0_{K_t} fx2n+1(t)=0Kt 可推出 f x 2 n ( t ) ≠ 0 K t f_{x_{2n}}(t) \neq 0_{K_t} fx2n(t)=0Kt；
If x 2 n ≪ x 2 n + 1 x_{2n} \ll x_{2n+1} x2n≪x2n+1, then f x 2 n ≪ F f x 2 n + 1 f_{x_{2n}} \ll_F f_{x_{2n+1}} fx2n≪Ffx2n+1 (by isomorphism). Let t 0 t_0 t0 be the T T T-minimal element of { t ∣ f x 2 n + 1 ( t ) ≠ 0 K t } \{t \mid f_{x_{2n+1}}(t) \neq 0_{K_t}\} {t∣fx2n+1(t)=0Kt}. Since K t 0 K_{t_0} Kt0 is Archimedean, f x 2 n ( t 0 ) = 0 K t f_{x_{2n}}(t_0) = 0_{K_t} fx2n(t0)=0Kt and f x 2 n + 1 ( t 0 ) ≠ 0 K t f_{x_{2n+1}}(t_0) \neq 0_{K_t} fx2n+1(t0)=0Kt, so the right-hand side fails.
若 x 2 n ≪ x 2 n + 1 x_{2n} \ll x_{2n+1} x2n≪x2n+1，则由同构可知 f x 2 n ≪ F f x 2 n + 1 f_{x_{2n}} \ll_F f_{x_{2n+1}} fx2n≪Ffx2n+1。设 t 0 t_0 t0 为 { t ∣ f x 2 n + 1 ( t ) ≠ 0 K t } \{t \mid f_{x_{2n+1}}(t) \neq 0_{K_t}\} {t∣fx2n+1(t)=0Kt} 的 T T T- 极小元，因 K t 0 K_{t_0} Kt0 是阿基米德群，故 f x 2 n ( t 0 ) = 0 K t f_{x_{2n}}(t_0) = 0_{K_t} fx2n(t0)=0Kt 且 f x 2 n + 1 ( t 0 ) ≠ 0 K t f_{x_{2n+1}}(t_0) \neq 0_{K_t} fx2n+1(t0)=0Kt，右侧条件不成立。

The right-hand side of the equivalence is a Π 1 0 \Pi^0_1 Π10 formula, and the left-hand side ( x 2 n ≈ x 2 n + 1 x_{2n} \approx x_{2n+1} x2n≈x2n+1) is a Σ 1 0 \Sigma^0_1 Σ10 formula. Thus the set { n ∣ x 2 n ≈ x 2 n + 1 } \{n \mid x_{2n} \approx x_{2n+1}\} {n∣x2n≈x2n+1} is Δ 1 0 \Delta^0_1 Δ10, so it exists by Δ 1 0 \Delta^0_1 Δ10 comprehension.

该等价式的右侧是 Π 1 0 \Pi^0_1 Π10 公式，左侧（ x 2 n ≈ x 2 n + 1 x_{2n} \approx x_{2n+1} x2n≈x2n+1）是 Σ 1 0 \Sigma^0_1 Σ10 公式，因此 { n ∣ x 2 n ≈ x 2 n + 1 } \{n \mid x_{2n} \approx x_{2n+1}\} {n∣x2n≈x2n+1} 是 Δ 1 0 \Delta^0_1 Δ10 集合，其存在性由 Δ 1 0 \Delta^0_1 Δ10 概括公理保证。

By Theorem III.2, this set is exactly the range of f f f, so arithmetical comprehension holds (Lemma I.4), i.e., ACA 0 \text {ACA}_0 ACA0 is proven.
□ \square □

由定理 3.2 可知，该集合恰好是 f f f 的值域，故算术概括公理成立（引理 1.4），即 ACA 0 \text {ACA}_0 ACA0 得证。
□ \square □

Corollary to Main Theorem III

主定理 3 的推论

There exists a computable ordered abelian group G G G such that there is no computable embedding of G G G into a Hahn subgroup.

存在可计算的有序交换群 G G G，其不存在到任何哈恩子群的可计算嵌入。

Proof Let REC \text {REC} REC be the model of RCA 0 \text {RCA}_0 RCA0 consisting of the natural numbers and recursive sets. REC \text {REC} REC does not satisfy arithmetical comprehension (since there exist recursive functions whose ranges are recursively enumerable but not recursive---by Lemma I.4, arithmetical comprehension fails in REC \text {REC} REC).
证明设 REC \text {REC} REC 为 RCA 0 \text {RCA}_0 RCA0 的模型，其元素为自然数和递归集。 REC \text {REC} REC 不满足算术概括公理（因存在递归函数，其值域是递归可枚举但非递归的集合 ------ 由引理 1.4，算术概括公理在 REC \text {REC} REC 中不成立）。

Let f f f be a recursive function with a non-recursive range. Construct G G G from f f f as in Theorem III.2--- G G G is computable (since its existence is provable in RCA 0 \text {RCA}_0 RCA0, and REC \text {REC} REC is a model of RCA 0 \text {RCA}_0 RCA0).

设 f f f 为值域非递归的递归函数，按定理 3.2 的构造由 f f f 生成有序交换群 G G G------ G G G 是可计算的（因其存在性可在 RCA 0 \text {RCA}_0 RCA0 中证明，且 REC \text {REC} REC 是 RCA 0 \text {RCA}_0 RCA0 的模型）。

If there existed a computable embedding of G G G into a Hahn subgroup, then the range of f f f would be recursive (by the proof of Main Theorem III), contradicting the choice of f f f. Thus no such computable embedding exists.
□ \square □

若 G G G 存在到某哈恩子群的可计算嵌入，则由主定理 3 的证明可知 f f f 的值域是递归的，与 f f f 的选择矛盾。因此，不存在这样的可计算嵌入。
□ \square □

Note

注

Hahn's embedding theorem is also provable in ACA 0 \text {ACA}_0 ACA0, hence it is equivalent to ACA 0 \text {ACA}_0 ACA0 over RCA 0 \text {RCA}_0 RCA0. The proof of Hahn's embedding theorem in ACA 0 \text {ACA}_0 ACA0 is a modification of the standard proof (see [11]) to avoid induction on non-arithmetical formulae, and can be found in [8].

哈恩嵌入定理同样可在 ACA 0 \text {ACA}_0 ACA0 中证明，因此在 RCA 0 \text {RCA}_0 RCA0 上，哈恩嵌入定理与 ACA 0 \text {ACA}_0 ACA0 等价。 ACA 0 \text {ACA}_0 ACA0 中哈恩嵌入定理的证明是对经典证明（参见文献 [11]）的修改，避免了对非算术公式的归纳，具体证明可参见文献 [8]。

BIBLIOGRAPHY

参考文献