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It's well-known that in Tao's Analysis I P28, he provides a provement of Trichotomy of order for natural numbers as follows.

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Denote the number of correct propositions among the three (i.e. $a<b,\ a=b,\ a>b$) by $\alpha$. First he shows $\alpha\not>1$, then proves $\alpha\ge 1$, then he concludes that $\alpha=1$. But this reasoning requires the proposition that we are proving. Only $\alpha\not>1\Rightarrow\alpha\le 1$ can we apply the anti-symmetric of order to conclude that $\alpha=1$.

My question is does Tao's provement formulate a circular argument？Is there other explanation s.t. $\alpha\not>1\wedge\alpha\ge 1\Rightarrow\alpha=1$ without $\alpha\le 1$ holding to bypass the application of this to-be -proved proposition ?

update:In comment area: "What do mean by 'But this reasoning requires the proposition'? – balddraz" My reply: I mean that he proves $\alpha=1$ by $\alpha\not>1\wedge\alpha\ge 1$ but this reasoning seems to require $\alpha\not>1\Rightarrow \alpha\le 1$ (which is what the trichotomy of order for natural numbers implies ) and then apply the anti-symmetric to complete the prove.

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  • $\begingroup$ This is a standard proof by induction, which may appear to circular until you become comfortable with the process. We show that the proposition is true in some base case. We assume it is true in some general case, and we show that when it is true, it is true when we increment by 1. If it is true in the case a=0, then it is true in the case a=1. If it is true in the case a=1 then it is true in the case a=2, etc. $\endgroup$
    – user317176
    Commented Apr 16 at 8:02
  • $\begingroup$ What do mean by "But this reasoning requires the proposition"? What part of the reasoning seems circular to you? $\endgroup$
    – balddraz
    Commented Apr 16 at 8:05
  • $\begingroup$ The author proves that exactly one case holds, first showing that at most one of the three cases holds and then showing that at least one of the three cases holds. This presupposes that we are able "to count" in the current mathematical environment were we are discussing the properties of natural numbers. $\endgroup$
    – Mauro ALLEGRANZA
    Commented Apr 16 at 8:08
  • $\begingroup$ So, did you mean that the "to count" system is independent of the natural system based the the peano's axioms ? And Can I hold the view that the chaper 2 of tao's analysis I is proving the two system are coinciding ? $\endgroup$
    – Richard Mahler
    Commented Apr 16 at 8:22
  • $\begingroup$ In a nutshell: YES; when and where did we learned how to count? Using Peano's axioms? $\endgroup$
    – Mauro ALLEGRANZA
    Commented Apr 16 at 8:30

2 Answers 2

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I believe your question is only regarding the trichotomy of order with $b=1$ which Tao uses to prove the proposition. That is, we need to rigorously prove that $a<1$, $a=1$ or $a>1$. Note that here we only need to show that at least one of these is true.

Case $1: a=0$

Clearly $a<1$ by definition.

Case $2:$ $a=n++$ for some $n\in \mathbb{N}$.

Subcase $1: n=0$

Here we get $a=1$ is true.

Subcase $2: n=m++$ for some $m\in \mathbb{N}$.

Then $a=n+1$ and $a\neq 1$ gives $a>1$.

Note that this sort of casework is, in principle, possible for any natural number $b$ instead of $1$. But rigorous proof requires induction on $a$ as is done in Tao's 'Analysis'.

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After discuss with my fellow, I recognized that we only need to show that $\alpha\not>1\Rightarrow \alpha\le 1$ s.t. the provement be completed.

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