I am new to analysis and started reading a PDF I found on Reddit, the link is here.
I stumbled on a few question about basic Peano axioms and the definitions that the PDF derived from it.
In case someone do not want to open the PDF, I am going to quote the first axiom and the second axiom from the book and ask my question.
In Peano arithmetic, we assume we have a set $\mathbb{N}$ (the natural numbers). We assume given $0 \notin \mathbb{N}$, and form $\tilde{\mathbb{N}} = \mathbb{N} \cup \{0\}$. We assume there is a map $$ s: \tilde{\mathbb{N}} \rightarrow \mathbb{N} $$ which is bijective. That is to say, for each $k \in \mathbb{N}$, there is a $j \in \mathbb{\tilde{N}}$ such that $s(j) = k$, so $s$ is surjective; and furthermore, if $s(j) = s(j′)$ then $j = j′$, so $s$ is injective. The map $s$ plays the role of “addition by 1,” as we will see below. The only other axiom of Peano arithmetic is that the principle of mathematical induction holds. In other words, if $S \subset \tilde{\mathbb{N}}$ is a set with the properties $$ 0 \in S, k \in S ⇒ s(k) \in S $$ then $S = \tilde{\mathbb{N}}$. Actually, applying the induction principle to $S = \{0\}\cup s(\tilde{\mathbb{N}})$, we see that it suffices to assume that $s$ is injective; the induction principle ensures that it is surjective.
It is not obvious to me how the application of induction principle on $S$, and assuming that $s$ is injective (one-to-one) can tell us that $s$ is also surjective (onto).
I assumed that the proof might start by proving that if $k \in \mathbb{N}$ and there is a $j$ such that $s(j) = k$, then there is also a $j'$ such that $j \neq j'$ and $s(j') = k + 1$, but I am not sure how to prove that.
Also, in the next paragraph it defines,
We define addition $x + y$, for $x, y \in \tilde{\mathbb{N}}$, inductively on $y$, by $x + 0 = x, x + s(y) = s(x + y)$.
I have two questions here, 1. what does it mean it to define $x + y$ inductively on $y$, and 2. how does proving $x + s(y) = s(x + y)$ defines $x + y$?
Edit 1: From reading the PDF a little bit more, I think I have an assumption that what it might mean to define inductively. My intuition is that by saying define inductively, the author might have meant to define something recursively. For example, definition of $x + y$ is recursively as follows, $x + 0 = x$, when $y = 0$, (base case) and, $x + s(y) = s(x + y)$, which can be interpreted as recursive definition, for to get the sum of $x$ and the "next step" of $y$, ($s(y)$), get the "next step" of $x+y$, which is $s(x + y)$. (I am not sure if this is what is supposed to mean, or if this is correct).
