鲁宾逊基本定理的现实意义

   六十年前，伟大的数学家鲁宾逊第一次证明了两条纯粹数学定理，名垂千古。

  去年11月，国家教育部发文，要求全国普通高校以及相关研究部门设立基础数学中心（也就是纯粹数学中心）.

  基础数学中心离不开鲁宾逊基本定理。

  鲁宾逊两条基本定理是什么？请见本文附件。

  国家的需求就是鲁宾逊基本定理的现实意义。

袁萌 陈启清 1月3日

附件：

Fundamentals of Model

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Theorem 10

yRobinson nally solved the centuries old problem of innitesimals in the foundations of calculus. Theorem 10. (The Leibniz Principle) There is an ordered eld ∗R called the hyperreals, containing the reals R and a number larger than any real number such that any statement about the reals which holds in R also holds in ∗R.

Proof. Let R be hR,+ + +,· · ·,< < <,0 0 0,1 1 1i. We will make the statement of the theoremprecise by proving that there is some model H, in the same language L as R andwith the universe called ∗R , such that R H and there is b ∈∗R such that a < b for each a ∈R. For each real number a, we introduce a new constant symbol ca. In addition, another new constant symbol d is introduced. Let Σ be the set of sentences in the expanded language given by: ThRR∪{ca < d : a is a real} We can obtain a model C |= Σ by the compactness theorem. Let C0 be the reduct of C to L. By the elementary diagram lemma R is elementarily embedded in C0, and so there is a model H for L such that C0 ∼ = H and R H. Take b to be theinterpretation of d in H.

3. DIAGRAMS AND EMBEDDINGS 28

theories in L1 and L2 respectively. Then T1∪T2 is satisable i there is no sentence σ of L such that T1 |= σ and T2 |= ¬σ. Proof. The direction ⇒ is easy and motivates the whole theorem. We begin the proof in the ⇐ direction. Our goal is to show that T1 ∪T2 is satisable. The following claim is a rst step. Claim. T1 ∪{ sentences σ of L : T2 |= σ} is satisable. Proof of Claim. Using the compactness theorem and considering conjunctions, it suces to show that if T1 |= σ1 and T2 |= σ2 with σ2 a sentence of L, then {σ1,σ2}is satisable. But this is true, since otherwise we would have σ1 |= ¬σ2 and hence T1 |= ¬σ2 and so ¬σ2 would be a sentence of L contradicting our hypothesis. This proves the claim.

The basic idea of the proof from now on is as follows. In order to construct a model of T1 ∪T2 we construct models A |= T1 and B |= T2 and an isomorphism f : A|L → B|L between the reducts of A and B to the language L, witnessing that A|L∼ = B|L. We then use f to carry over interpretations of symbols in L1 \Lfrom A to B , giving an expansion B∗ of B to the language L1 ∪L2. Then, sinceB ∗|L1 ∼ = A and B∗|L2 = B we get B∗ |= T1 ∪T2. The remainder of the proof will be devoted to constructing such an A, B and f. A and B will be constructed as unions of elementary chains of An’s and Bn’s while f will be the union of fn : An ,→ Bn. We begin with n = 0, the rst link in the elementary chain. Claim. There are models A0 |= T1 and B0 |= T2 with an elementary embeddingf 0 : A0|L ,→ B0|L. Proof of Claim. Using the previous claim, let A0 |= T1 ∪{ sentences σ of L : T2 |= σ} We rst wish to show that Th(A0|L)A0∪T2 is satisable. Using the compactness theorem, it suces to prove that if σ ∈ Th(A0|L)A0 then T2 ∪{σ} is satisable. For such a σ let ca0,...,can be all the constant symbols from LA0 \L which appear in σ. Let be the formula of L obtained by replacing each constant symbol cai by a new variable ui. We have A0|L|= [a0,...,an] and so A0|L|= ∃u0 ...∃un By the denition of A0, it cannot happen that T2 |= ¬∃u0 ...∃un and so there is some model D for L2 such that D |= T2 and D |= ∃u0 ...∃un. So there are elements d0,...,dn of D such that D |= [do,...,dn]. Expand D to a model D∗ for L2 ∪LA0, making sure to interpret each cai as di. Then D∗ |= σ, and so D∗ |= T2 ∪{σ}. Let B∗ 0 |= Th(A0|L)A0∪T2. Let B0 be the reduct of B∗ 0 toL2; clearly B0 |= T2.Since B0|L can be expanded to a model of Th(A0|L)A0, the Elementary Diagram Lemma gives an elementary embedding f0 : A0|L ,→ B0|L and nishes the proof of the claim.

3.