Peano axioms-Analysis I by Terence Tao

Question

Statement: Terence Tao in his book Analysis I states that the set N = {0,0.5,1,1.5,2,...} satisfies peano axioms 1 to 4. Axiom 2: if n is a natural number, n++ is also a natural number Definition 2.1.3: 0++ =1, 1++ = 2,2++ = 3 my understanding : because of definition 2.1.3, the set N in the statement doesn’t satisfy axiom 2. My question : why did Tao state that set N satisfies all 4 axioms. What maybe the reasoning behind it?

Answer 1

On pages 17 and 18 Tao casually (non)defines $n++$ as "the successor" which means counting forward. $0++$ counting forward is $1$ and $2++$ counting forward is $3$ and so on. (This is an intuitive definition for the sake of making what is a purely abstract concept to mathematician, a solid example to a novice student.)

As such if we were to talk about the successor of non-integers $x++$ would be what we think of as $x+1$. So $\pi++ = \pi + 1 = 4.141592653......$.

Now his idea, (which I guess is a good one) is rather than introduce the natural numbers purely by abstract Peano postulates where we say "$\mathbb N$ is a set. We don't anything about it except it obeys the axioms: $0$ is a in $\mathbb N$. We don't know what $0$ is. It's just a thing. Every thing $n$ in the set has the concept of there being another element $n++$. We don't know what that means that just a concept, but ...". Instead Tao seems to say "Look, we all know that $\mathbb N = \{0,1,2,3,.....\}$ and that we or just counting. But what does that mean? How can we formalize that."

Okay. Suppose we have two sets $\mathbb N = \{0,1,2,3,4,5,6,......\}$ and $N'= \{0, 0.5, 1, 1.5, 2, 2.5, 3.0, 3.5, .....\}$.

Axiom 1. $0 \in \mathbb N$. That's clearly true for how we listed $\mathbb N$. And for our made up set $N'$ it appears also to be true.

Axiom 2. If $n\in \mathbb N$ then $n++\in \mathbb N$. That is clearly true for $\mathbb N$. $0\in \mathbb N$ and $0++ = 1 \in \mathbb N$ and $1\in \mathbb N$ and $1++ =2 \in \mathbb N$ and so on.

AND IT IS ALSO TRUE FOR N'. $0 \in N'$ and $0++=1\in N'$. And $0.5\in N'$ and $0.5++ = 1.5\in N'*$ and so on.

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so that's your answer

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It's also clear that Axioms 3 and 4 hold.

Axiom 3: $0$ is not equal to any $n \in \mathbb N$.

It is also true that $0$ is not a successor to any element of $N'$. As long as we are cheating by assuming we know things about numbers (that is very unorthodox of Tao to do so, but I must admit it probably is much easier on the novice student) we know $(-1)++ = 0$ and $-1 \not \in \mathbb N$ and $-1 \not \in N'$.

Axiom 4: Different elements of $\mathbb N$ must have different successors. If $m\ne n$ then $m++ \ne n++$. And if $m++ \ne n++$ then $m \ne n$.

That's true for both cases.

As all true for both $\mathbb N$ and $N'$ (and for $\mathbb Q^+$ or $\mathbb R^+$). In fact these axioms are true for any sets that have $0,1,2,3,4,.....$ and any other extra values so long as if it has $x$ it also has $x+1, x+2, x+3,....$ and so long as it doesn't have $-1$.

So what allows us to say $\mathbb N$ is just $0,1,2,3,4,.... $ and not any of these other sets?

That'd be axiom 5. The induction axiom.

If $P(0)$ is true. And $P(n)$ implies $P(n+1)$ [which means $P(0), P(1), P(2), P(3)..... $ etc. must all be true] then we have to have it true for all elements in the set.

This fails for $N'$ as, we could conclude $P(0), P(1), P(2)...$ etc or true we have no reason to thing $P(0.5)$ is true. This axiom basically says, $\mathbb N$ contains only the terms we can count to from $0$ and nothing more. We can count to the half integers so the half integers are not part of the natural numbers.