How can initial segment of the set of all natural numbers is equal to itself

Question

I'm studying ordinal numbers in Halmos's book (page 75). After the definition of ordinal number, he proves that $w^+$ is an ordinal number where $w$ is the set of all natural numbers.

In the last part of the proof, he states that $s(w) = w$ "by definition of order" (where $s(w)$ is the initial segment of $w$). (I don't actually know which definition has been applied in this context)

I couldn't understand how $s(w) = w$ because as far as I know, we don't have the notion of $x \lt w$ where $x$ is a natural numbers (i think we only have the order defined on the set of natural number, in other words, we compare 2 natural numbers but not a natural number and the set of all natural numbers). And moreover, I think even if we can "compare" a natural number and $w$, how could we make sure that the initial segment of $w$ is equal to $w$ ? (i.e. I think there can be for example 2 times the number 1 appear in the initial segment i.e. $s(w) =$ {0; 1; 1; ...}).

Could you please help me out of this misunderstanding ? Thank you very much

An example of an ordinal number that is not a natural number is the set $w$ consisting of all the natural numbers. This means that we can ready "count" farther than we could before; whereas before the only numbers at our disposal were the elements of $w$, now we have $w$ itself.

We have also the successor $w^+$ of $w$; this set is ordered in the obvious way, and, moreover the obvious ordering is a well ordering that satisfies the condition imposed on ordinal numbers.

Indeed, if $x \in w^+$, then by the definition of successor, either $x \in w$ in which case we already know that $s(x) = x$ or else $x = w$, in which case $s(x) = w$, by the definition of order, so that again $s(x) = x$

Answer 1

Halmos defines an ordinal number in the paragraph before the ones you are quoting:

The chief application of the axiom of substitution is in extending the process of counting far beyond the natural numbers. From the present point of view, the crucial property of a natural number is that it is a well ordered set such that the initial segment determined by each element is equal to that element. (Recall that if $m$ and $n$ are natural numbers, then $m < n$ means $m\in n$; this implies that $\{m \in w: m < n\} = n$.) This is the property on which the extended counting process is based; the fundamental definition in this circle of ideas is due to von Neumann. An ordinal number is defined as a well ordered set a such that $s(x) = x$ for all $x$ in $\alpha$; here $s(x)$ is, as before, the initial segment $\{\eta \in\alpha: \eta < x\}$.

Since the initial segment of $x$ is $x$ itself, we must have that $y\in s(x)$ (which is the same as $y<x$) if and only if $y\in x$; in other words, the ordering of an ordinal set is the same as $\in$.

When Halmos says that $\omega^+$ is ordered in the obvius way, he means that elements of $\omega$ are ordered in the common way and $\omega$ is bigger than all of them (you can easily check that this is a well order), which matches with $m<n$ if and only if $m\in n$, because we already know this is true for $m, n\in\omega$, and we have $m<\omega$ for all $m\in\omega^+$ different from $\omega$, which are precisely the $m\in\omega$.