Understanding the natural numbers and Peano's axioms I am reading the book Analysis $I$ by Terence Tao, where the following is written: [...] $\mathbf{N}$ should consist of $0$ and everything which can be obtained from incrementing: $\mathbf{N}$ should consist of the objects $0,0++,(0++)++,...$ (with $++$ he denotes the increment/successor function). He then comes to
Axiom 2.1 $0$ is a natural number.
Axiom 2.2 If $n$ is a natural number, then $n++$ is a natural number.
Later (on page 21) he writes: Assumption There exists a number system $\mathbf{N}$ whose elements we will call the natural numbers for which the Axioms $2.1-2.5$ (the Peano-Axioms) are true.
$(Q1)$ What does this mean? Does this mean that we assume that there is some system $\mathbf{N}$ that denotes natural numbers, where we don't know what exactly they are. In the context of set theory we thus have a set $\mathbf{N}$ whose objects are called natural numbers (whatever objects are; I have not seen any definition of this). Thus its objects could be anything in the beginning. Axiom $2.1$ now says that $0$ is a natural number. This should mean that $\mathbf{N}$ contains at least one object and one of those will be called $0$. We couldve however chosen any name or symbol here, as far as I know, such as for example $A$ or $a$ or even $5$. Now Axiom $2.2$ states that the set contains more than one element, namely infinitely many elements and the increment of the previous element will be called $n++$. Again the symbols here should be arbitrary and could be anything.
He now proceeds to define $1:=0++,2:=1++,3:=2++$ and so on. What this now does is it defines the underlying object of the symbol $1$ to be the same as the underlying object of $0++$ and so on. Thus the object that $1$ stands for is now a natural number. So what we have done is basically given the objects "simpler" (in terms of notation) names.
I could've also called the zero object $A$, $B:=0++$, $C:=B++$,... giving the objects different names. (Here I mean that this should also be done in a "good" way, such that essentially $A$ represents $0$, $B$ is $1$ and so on such that $J$ is $9$ and then $BA$ is $10$ and so on.
$(Q2)$ If this is the case, what is the difference between the natural numbers presented "commonly" by the symbols $0,1,2,3,...$ and the ones I defined by $A,B,C,...?$ Essentially they should be the same set, however, no one would say that the set $\{0,1,2,3,...,10,11,12,...\}$ and $\{A,B,C,...,BA,BB,BC,...\}$ are equal would they? Maybe I am wrong here, but are they actually the same set, because I defined them to mean the same underlying object, such that the symbols don't make a difference? Basically what confuses me is if the symbols matter, meaning that they make a difference, even when defined to mean the same object.
I think the above set should, in this context of defining them what they mean be literally equal. However, if I consider the set $\{0,I,II,III,IV,V,...,X,XI,...\}$ without defining anything, then $\{0,1,2,3,4,...\}$ should only be bijective to this set and not literally equal, right? This would mean that the difference is the context, in which I have defined the letters to be the objects that the symbols $1,2,3,...$ represent. In the book Tao says, that actually $\{0,I,II,III,IV,...,\}$ would be different from $\{0,1,2,3,...\}$, if one "wanted to be annoying" (guess I am kinda pedantic), despite them both referring to the natural numbers. How can this be? Do symbols make a difference, despite meaning the same underlying object?
Where I am coming from: I am basically confused about the concept of sets and "objects" and tried to give some "extreme" example in order to make sure I understand it correctly. As far as I understand, a set can contain objects, that I can't write down, which is why symbols are used. Different symbols can mean the same thing, however, which is why the context matters, as presented in the examples I gave. What caused this was my question whether $e^{2 \pi i} \in \{a+ib \ | a,b \in \mathbf{R}\}$ despite it not being a "fitting" symbol. I hope this helps making my question more clear.
 A: You are raising legitimate questions, but you are also conflating several different questions.
First, there is the issue of there being multiple designations for the same thing.
Secondly, there is the issue of "what natural numbers really are".
I will focus solely on the second question for now.
The answer is that it doesn't matter at all what the natural numbers really are. The question is totally irrelevant (though it is not stupid).
But what does matter is what properties the natural numbers have. In other words, what matters is the structure of the set of natural numbers, considered as a whole, and how this structure relates to other mathematical structures.
We say that a set $\mathbb{N}$, together with an element $0 \in \mathbb{N}$ and a function $++ : \mathbb{N} \to \mathbb{N}$, is "a set of natural numbers" if all of the properties that Terence Tao lays out are satisfied by $\mathbb{N}, 0, ++$.
We are glossing over some more foundational questions - what is a set, and what is a function? But let's ignore those, since we'll get too bogged down.
The sort of thing that the actual elements of $\mathbb{N}$ are is not part of the definition of "a set of natural numbers". In fact, when one is doing mathematics, it's fairly common to spot all kinds of sets (together with a distinguished element and function) which match the definition of "a set of natural numbers".
The key thing here is that any two "sets of natural numbers" are equivalent in exactly 1 way. In particular, given two "sets of natural numbers" $(\mathbb{N}, 0, ++)$, $(\mathbb{N}', 0', ++')$, there is a unique bijection $f : \mathbb{N} \to \mathbb{N}'$ such that $f(0) = 0'$ and for all $n \in \mathbb{N}$, $f(n++) = f(n)++'$.
This basically means that once you know that something is "a set of natural numbers", you have specified all the structural properties of this set. This is why we often refer to the set of natural numbers, as there is essentially only one up to bijection.
The "axiom of infinity" is the statement "There is some $(\mathbb{N}, 0, ++)$ which is a set of natural numbers".
One question one might ask is whether $(\mathbb{N}, 0, ++)$ and $(\mathbb{N}', 0', ++')$ are "equal". There are actually three different answers to this question, depending on what foundational system you're using.
Answer 1: It makes sense to ask this question, and the answer is almost always "no". This is the answer we get when we are doing mathematics under the most commonly used foundation of mathematics, which is "material set theory" (most commonly given by the axioms of ZFC). This is because material set theory actually has a very strong notion of "what things really are", although it's a notion which is rather subtle and difficult for newcomers to understand and which is totally irrelevant for doing most of mathematics. This means that $(\mathbb{N}, 0, ++)$ actually contains all kinds of "extra data" which is not relevant to the "set of natural numbers" structure, and this "extra data" may not be the same as the "extra data" that $(\mathbb{N}', 0', ++')$ comes with.
Answer 2: It does not even make sense to ask whether $(\mathbb{N}, 0, ++)$ and $(\mathbb{N}', 0', ++')$ are equal. In fact, we should not even be permitted to raise such a question. This is the answer given by so-called "structural set theory" (most commonly given by the axioms of ETCS, which uses category theory). Under this theory, there is a basic notion of equality for two elements $a, b \in C$ of the same set, so it makes sense to ask whether $a = b$ where $a$ and $b$ are both elements of $C$. There is also a basic notion of equality for two functions $f, g : A \to B$ which have the same domain and codomain, so it makes sense to ask whether $f = g$. But what does not make sense is to ask whether two sets are equal, whether elements of different sets are equal, and whether two functions that don't share the same domain and codomain are equal. The only thing we can say is that the two structures are isomorphic.
Answer three: it does make sense to ask the question, and the answer is always "yes". This is the view of the relatively new foundational theory known as Homotopy Type Theory. In homotopy type theory, there is a strong notion of "equivalence" between any two structures of the same kind, and there is an axiom that "equivalence is equivalent to equality". In particular, the fact that there is a bijection $f : \mathbb{N} \to \mathbb{N}'$ which preserves the full structure $(\mathbb{N}, 0, ++)$ means that $(\mathbb{N}, 0, ++) = (\mathbb{N}', 0', ++')$. This foundational theory comes will all kinds of subtleties - for example, two things can be equal to each other in more than one way. However, in the case that we're comparing two sets of natural numbers $(\mathbb{N}, 0, ++)$ and $(\mathbb{N}', 0', ++')$, there is only one bijection between them that preserves their structure; they are therefore equivalent in only one way, and hence equal in only one way. It takes quite a while to fully understand what's going on here, so I wouldn't worry too much about it unless you decide it's an interesting enough idea to be worth studying in a few years.
The bottom line is this: when you are actually doing mathematics, the only thing that matters is that there is some $(\mathbb{N}, 0, ++)$ which satisfies the axioms of a set of natural numbers. The more complicated questions about the "true nature" of the natural numbers have wildly different answers depending on which foundational theory you're using, but in practice, what matters is the Peano axioms.
