In Terence Tao's Analyis Volume One. He presents that following Lemma. I would like to know if the argument presented below is correct? For your ease i have presented the relevant Peano's axiom as presented in Tao's Volume.

Axiom 2.4. Different natural numbers must have different successors; i.e., if $n, m$ are natural numbers and $n \neq m$, then $n++ \neq m++$. Equivalently, if $n++ = m++$, then we must have $n = m$.

Lemma $2.2.10$. Let $a$ be a positive number. Then there exists exactly one natural number $b$ such that $b\text{++} =a$.

Prior to proceeding with the argument let us first reformulate the statement of Lemma $\textbf{2.2.10}$ to say that. Given any natural number $a$ if $a$ is positive then there exists a unique natural number $b$ such that $b\text{++} = a$.

Proof. We construct the proof by recourse to mathematical Induction on $a$. For $a = 0$ the statement in question is vacuously true since $a$ is $0$ and therefore not positive.

Now assume that the claim in question is true for an arbitrary natural number $k$. Consider first the case where $k=0$. then choosing $b = 0$ would imply that $b\text{++}= 1$ as required. In addition if $\alpha\text{++}= 1$ for some arbitrary natural number $\alpha$ then axiom $\textbf{2.4}$ would imply $\alpha = b$ demonstrating the uniqueness of $b$.

Now consider the case where $k$ is positive. Since $k$ is positive using the inductive hypothesis we may choose a unique $\beta$ such that $\beta\text{++} = k$. Now assume that $k++$ is positive and choose $b = k$ evidently $b++ = k++$ as required. Now assume for an arbitrary natural number $\alpha$ that $\alpha++ = k++$ as in the previous case use of axiom $\textbf{2.4}$ yields $\alpha = k = b$. This completes the inductive step.


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    $\begingroup$ Does Tao really use $++$ to signify successor? You usually don't see it in mathematics. $\endgroup$ – Arthur Apr 1 '18 at 19:57
  • $\begingroup$ @Arthur Yes it does represent the successor $\endgroup$ – Atif Farooq Apr 1 '18 at 19:58

All good, but the inductive step can be done a bit simpler:

All we need to show is that there is a unique $a$ such that $a++=k++$. Well, clearly this is true for $a=k$, and so there is at least one, and for any $b$ for which $b++=k++$, we have by axiom 2.4 that $b=a$, and so this $a(=k)$ is unique.

In other words, we don't use the inductive hypothesis, and so don't even need to spell it out. As such, you also don't need to separate the '$k=0$' case from the '$k$ is positive' case.


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