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第一个补充：
细心的读者也许会发现，《陶哲轩实分析》第97页给出的关于命题5.5.12的证明是有点“不严格”的。为什么呢？因为该页第9行中的“可以取0＜ε＜1使得x^2+5ε＜2”以及第13、14行中的“我们可以取一个ε，0＜ε＜1使得x^2-4ε＞2”并没有提及怎么构造合适的ε。当然上面所需要的两个合适的ε都很容易构造出来，以致省略也并不影响严谨性。接着看第101页习题5.6.1中的“提示：复习命题5.5.12的证明”，它实质上提示的是引理5.6.6(a)的证明。而引理5.6.6(a)的证明同样涉及如何构造合适的ε的问题。为此，我曾请教过陶哲轩如何把相应的ε构造出来。但他只是说这样的ε总是存在的，只要进行足够多的计算。以下是他的回复：
Try working with a specific numerical example, e.g. show that there exists y such that y^3 = 5, to build intuition. The point is that one does not need to choose epsilon right away,but only after one performs enough computations to simplify one's upper and lower bounds on such expressions as (y-epsilon)^3.
相对于严格的证明来说，陶的说法有点美中不足。另一方面，后来发现Walter Rudin的PMA教材（英文第3版，机械工业出版社）第10页中的h和k两个量正是两个合适的ε。换句话说，存在合适的ε的一般构造方法。遗憾的是，Rudin的证明需要用到b^n-a^n=(b-a)(b^(n-1)+b^(n-2)a+…+a^(n-1))这样的恒等式，而陶的书在引理5.6.6之前并未提及这个恒等式，所以不能直接用，否则就破坏了陶的教材中一贯的严格风格。所以要这个恒等式的话还得先对它进行证明。当然，我是后来才看到Rudin的证明，我也并没有采用他的方法。也许是巧合吧，我发现根据陶的书在引理5.6.6之前提及的定理和命题等可以严格证明引理5.6.6(a)。以下就是我写的证明（符号太多，只好以图片的形式上传；字不是正楷的，见谅）：
第二个补充：
《陶哲轩实分析》第17页第一段向读者解释了公理2.5的作用，但陶的解释恐怕还是让很多读者一头雾水，一名弗吉尼亚大学的学生就问过相关的问题。我当时觉得他问的问题有趣，几经思考，写了一个英文帖子来说明公理2.5的作用。很抱歉，当时写的就是英文，过后我也没想过要译成中文。并且我的论述也是非正式的，旨在比较直观地说明这个公理的作用。以下就是我写的那个帖子：
Consider the Axioms 2.1-2.4 of the Peano Axioms:
Axiom 2.1. 0 is a natural number.
Axiom 2.2. If n is a natural number, then n++ is also a natural number.
Axiom 2.3. 0 is not the successor of any natural number; i.e., we have n++ ≠ 0 for every natural number n.
Axiom 2.4. Different natural numbers must have different successors; i.e., if n and m are natural numbers and n ≠ m, then n++≠ m++. Equivalently, if n++ = m++, then we must have n = m.
Starting from “0” to create a series of natural numbers,we find any of such natural numbers can be represented by no more than three basic symbols, that is, “0”, “( )” and “++”. Nevertheless, we should admit that with the above four axioms, there still exist at least two possibilities to create natural numbers.
(i) One would not lead to an incompatibility through defining or assigning a symbol such as “0.5”, “1.1”, “⊙” which is distinct from any natural number denoted by “0”, “( )” and “++”to be a natural number. We can test whether the Axioms 2.1-2.4 still hold. For the Axiom 2.1, certainly it doesn’t exclude “0.5”, “1.1” or “⊙” to be a natural number. For the Axiom2.2, we recursively define 0.5++ is the successor of 0.5, (0.5++) ++is the successor of 0.5++, and so forth. Also, we can replace 0.5 with 1.1 or ⊙ for the same process. For the Axiom 2.3, evidently, we can have n++ ≠ 0 for every natural number n. Though it may not be necessary, we can also have n++ ≠ 0.5, orn++ ≠ 1.1, or n++ ≠ ⊙ for every natural number n. Bring in theAxiom 2.4, it would not cause any contradiction considering “0.5”, “1.1” or “⊙”and their corresponding successors as natural numbers. First, we actually have defined 0.5 ≠ 0 in the very beginning; Secondly, we can define 0.5++ ≠ 0++,(0.5++)++ ≠ (0++)++, etc. Thus, it still satisfies the requirement of the Axiom 2.4 that different natural numbers must have different successors. Similarly,we can define 1.1++ ≠ 0++, (1.1++)++ ≠ (0++)++, …; we can also define ⊙++ ≠ 0++, (⊙++)++ ≠ (0++)++, … . Hence, no contradiction would happen even contain the Axiom 2.4. Further, we can define 0.5, 1.1 together with ⊙ to be three different natural numbers. Similar to the process shown above, we can define 1.1++ ≠ 0.5++, (1.1++)++≠ (0.5++)++, …, and ⊙++ ≠ 0.5++, (⊙++)++ ≠ (0.5++)++,…, and so forth.
(ii) One may define a different operation from increment to extend the extent of our natural number system. For instance, we can define a new operation such as “insert ⊙ forward”, then “⊙⊙”, “⊙⊙⊙”,…, are also natural numbers. Understandably, we define ⊙ ≠ ⊙⊙, ⊙⊙ ≠ ⊙⊙⊙, … .Through the same way shown in (i), we can create a new series of natural numbers from such as “⊙⊙”or “⊙⊙⊙”.
However, the Axiom 2.5 of the Peano Axioms excludes the above two possibilities. In another word, by Axiom 2.5, we can prove that every natural number can only be represented by no more than three basic symbols , “0”, “( )” and “++”.
Axiom 2.5. Let P(n) be any property pertaining to anatural number n. Suppose that P(0) is true, and suppose that whenever P(n) istrue, P(n++) is also true. Then P (n) is true for every natural number n.
注：以上证明和英文帖子均写2010年。此外，建议对经济问题感兴趣的读者看看我写的另外三篇文章，即《请以科学家的眼光看待经济问题》、Reconsider the Question That Why the Firm Exists 和《论非整体移动的需求曲线》。这三篇文章在论坛上都能找到，也可以用谷歌或百度直接搜索。肖演东，2012年2月2日。
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