0
$\begingroup$

I am failing to understand the intent of the question posed by exercise 3.5.13 of Tao's Analysis I 4th ed.

The purpose of this exercise is to show that there is essentially only one version of the natural number system in set theory (cf. the discussion in Remark 2.1.12). Suppose we have a set $N'$ of “alternative natural numbers”, an “alternative zero” $0'$ , and an “alternative increment operation” which takes any alternative natural number $n' \in N'$ and returns another alternative natural number $n'++' \in N'$, such that the Peano axioms (Axioms2.1-2.5) all hold with the natural numbers, zero, and increment replaced by their alternative counterparts.

Show that there exists a bijection $f : N \to N'$ from the natural numbers to the alternative natural numbers such that $f (0) = 0'$ , and such that for any $n \in N$ and $n' \in N'$, we have $f(n)=n'$ if and only if $f(n++)=n'++'$. (Hint: use Exercise 3.5.12)

I do understand that we are asked to find a function that is a bijection between $N$ and $N'$.

What I am unclear on is final phrase:

"and such that for any $n \in N$ and $n' \in N'$, we have $f(n)=n'$ if and only if $f(n++)=n'++'$."

Does this mean the function can be assumed to have the property that $f(n)=n' \iff f(n++)=n'++'$ ?

Or does it mean we have to show the bijection only holds if $f(n)=n' \iff f(n++)=n'++'$ ?

$\endgroup$
1
  • $\begingroup$ The bijection you're asked to find has to also satisfy that property, meaning that you have to prove it for whatever you're gonna find. Isn't it apparent? $\endgroup$
    – Vaskara_GRek_O
    Commented Jun 4 at 20:34

1 Answer 1

Reset to default
2
$\begingroup$

Neither. You have to show that $f$ satisfies two properties:

  • $f(0)=0'$

  • for any $n \in \mathbb{N}$ and ...

You also have to show that $f$ is a bijection.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .