Analysis I§
Abstract
这是陶哲轩先生赫赫有名的 实分析 I. 限于本人水平(不论是数学还是英语), 下面笔记中许多证明之严谨性惨不忍睹. 但是其欲图表述之含义应当是点到了. 在看原著和问 AI 都无效时看看下面的证明等等或许有所帮助.
1. "Natural Numbers"§
Note
Why do I put quotation marks around the "natural numbers"? Because it is not the same as natural numbers which we have learned in elementary school. As you will soon see, the natural numbers here is an abstract algebraic structure(e.g., the set of even numbers are perfectly satisfies the axioms which define natural numbers). However, this is not a point. It is just a matter of different encoding system.
1.1 The Peano axioms§
- (1) \(0\) is a natural number.
- (2) If a number is a natural number, its successor is a natural number(We denote successor operator as \(++\)).
- (3) \(0\) is not the successor of any natural number.
- (4) Different natural numbers must have different successors.
- (5) (Mathematical Induction) For any proposition concerning natural numbers, when: \(\text{BEGIN}\) i) it is true for \(0\); ii)if we suppose that a proposition is true for any \(n\), it is true for \(n++\) too. \(\text{END}\) It is true for all natural numbers.
The most incomprehensible and interesting axiom may be the fifth one: Mathematical Induction. However, it is also a significant axiom because it ensures that all natural numbers can be generated recursively starting from 0 using the successor operation.
Proof: Peano axioms (5) (Proposition \(I\)) \(\Longleftrightarrow\) All natural numbers can be generated recursively starting from 0 using the successor operation (Proposition \(M\))
Now we denote the proposition of number \(n\) as \(P(n)\).
(i) \(M\Longrightarrow I\): Let \(X\) satisfies Proposition \(M\) (\(x_0=0, x_1=x_0++,\dots\)). We need to prove that when the assumptions Proposition \(I\) are all satified the proposition is true for all natural numbers. The assumption 1 supposes that \(P(x_0)\) is true. Then, the assumption 2 supposes that for any \(n\) if \(x_n\) is true, and \(x_{n+1}\) is also true. Now \(P(x_0)\) is true, and because any number in \(X_n\) can be generated recursively starting from \(0\) using the successor operation (\(++\)), which means \(P(x_0++)=P(x_1)\) true, \(P(x_1++)=P(x_2)\) true \(\dots\) and thus \(P(x_n)\) is true for all \(x_n\) in \(X\).
(ii) \(I\Longrightarrow M\): (For better understanding, a contradiction proof is used here.) Suppose that \(M\) is false when \(I\) is true. The falsity of \(M\) implies that at least one element cannot be generated recursively starting from 0 using the successor operation. So let \(N=X\cup C\) where elements of \(X\) can be generated recursively starting from 0 using the successor operation and elements of \(C\) cannot. We suppose that \(P(x)\) is true while \(P(c)\) not. So \(P(n_0)\)(i.e. \(P(x_0)\)) is true, and \(P(n_0++)\)(i.e. \(P(x_0++)\)) \(=\) \(P(x_1)\), \(\dots\) According to the axiom 5, \(P(x)\) is true. However, \(P(c)\) is not true. Therefore, when \(I\) is true, \(M\) must be true.
In summary, Proposition \(I\) and Proposition \(M\) are equivalent. \(\Box\)
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