What are the first 3 ordinal numbers in the set of all integers?

Question

I've been reading a little bit about Cantor's work with transfinite numbers, and one point is confusing me.

If my understanding is correct, the set of positive integers ω and the set of negative integers ω* have the same cardinality (Aleph-0), but different order types. But it's simple for me to pull describe the first 3 ordinals in those sets because they have a single vector (incrementation or decrementation): [1,2,3,...] and [-1,-2,-3,...] This leads to confusion about how I describe the first 3 ordinals in the set of all integers because ℤ = [...,-3,-2,-1, 0, 1, 2, 3,...]

• What are the first three ordinal numbers in the set of all integers ℤ?

Answer 1

I'm really asking about how you can determine a starting point for natural ordinals in a set that doesn't have a beginning or ending.

I think you misunderstand what "ordinals" are. Ordinals do not generalize the integers; they generalize the natural numbers. Ordinals describe orderings with a certain important property: well-orderedness. A linear order is well-ordered if every (nonempty) subset of it has a least element. This is true of the natural numbers (think about proof by induction), but not of the integers (e.g. there is no least integer).

EDIT:

You write further

I'm trying to understand why ordinal numbers are restricted to a generalization of the natural numbers, and cannot also be a generalization of integers.

Well, there's no reason we couldn't have generalized the integers instead, but we chose to generalize the naturals. It comes down to what we want ordinals to do; and the motivation for ordinals is transfinite induction. Just like we have induction on the natural numbers, we can use induction on any well-ordering: if $(L,<)$ is a well-ordering, and $X\subseteq L$ is such that

• $X$ is nonempty, and

• If $x\in L$ and every $y<x$ is in $X$, then $x\in X$,

then $X=L$. (Why? Suppose otherwise, and think about the set $L\setminus X$ ...)

This is an incredibly useful technique in mathematics (as is one of its applications, definition by transfinite recursion). Ordinals are introduced as a way of codifying this technique.

Answer 2

You are misusing the word "ordinal number," which might indicate a confusion.

The ordinal numbers are $0,1,2,\dots.$ Historically, before some of modern set theory, ordinal numbers started at $1$, and could be thought of as a "place in a race" - first place, second place, etc. Starting at $0$ is nice for set theorists, and you can think of the sentence "Person $X$ is in the $n$th ordinal position" as meaning that there are exactly $n$ people who beat $X$ in the race. (But it gets more complicated when dealing with infinite ordinals.)

In general, a well-ordered set $X$ is a set with an order $\leq$ which has the property that if $S\subseteq X$ is non-empty, then $S$ has a least element.

If this were a race, the condition of well-ordered would mean that you can find one winner in any slice of the racers - you could take the racers who are left-handed botanists, and, if there were any left-handed botanists in the race, you could determine who won amongst the left-handed botanists.

Given a well-ordered set, you can say "Which element of $X$ is at position $n$?" where $n$ is an ordinal number. Sometimes, when $n$ is too large, the answer is none. For example, in a race of $4$ people, if we ask who came in fifth place (ordinal place $4$), you'd say "nobody."

So, your $\omega^*$ is well-ordered, as is $\omega$. So you can say "what are the first $3$ places in these orders", or you can think of it as "What races beat the racer in third place?"

But $\mathbb Z$ is not a well-ordered set. In particular, it doesn't have a least element. But $\mathbb Q^{\geq 0}$ is also not a well-ordered, even though $0$ is the least element of $\mathbb Q$, because there are subsets of $\mathbb Q^{\geq 0}$ which do not have a least element. We can't find a winner for all subsets of the "racers" in this race.

Answer 3

An ordinal number is to describe the order of a number within a well-ordered set. A well ordered set is a set of elements and an order so that certain properties are met[$*$].

The Natural numbers with the order "$>$" being the usual order we've known and loved since kindergarden is a well-ordering.

The negative integers with the order "$>$" is not a well-ordering. However if we use the order "$<$", it is. (And in turn the Natural numbers with the order "$<$" is not well ordered.)

The integers are not well-ordered with either "$<$" or with "$>$". However "$<$" and "$>$" are not even remotely close to the only orders.

Let $>_*$ be defined as $a >_* 5$ for all $a \in Z$. Otherwise if $a< 0 < b$ then $b >1> a$. And if neither of those apply, then $b >_* a \iff |b| > |a|$. That is well ordered and the first three ordinals are $5, -1, -2$.

But that's a silly and arbitrary ordering.

If two sets $X$ and $Y$ have the same cardinality, then there exists a bijection, $f:X \rightarrow Y$. If $X$ has a well-ordering $>_X$ then defining an order $>_Y$ on $Y$ as $a>_Y b \iff f^{-1}(a) >_X f^{-1}(b)$ will be a well-ordering.

So for example if $f:\mathbb N\rightarrow \mathbb Z$ via $f(n) = \frac n2$ if $n$ is even, and $f(n) = \frac {n+1}2$ if $n$ is odd, then the order "$>_Z$" defined as $z >_Z y \iff f^{-1}(z) > f^{-1}(y)$ we be a well-ordering; we will have $0 <_Z -1 <_Z 1 <_Z -2 <_Z 2 ....$; and the "first three" integers will be $0,-1,1$.

And just for giggles. Remember $>_*$ above? The weird one? We can use that to define $>_N$ on $\mathbb N$ where $a >_N b \iff f(a) >_* f(b)$. That would order the Natural numbers as $\{10, 1,3,5,7....\}$.

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[$*$] 1) exactly one of $a >> b; a=b; b>0$ must be true. 2) $a >>b; b>>c \implies a>>c$. And 3) all subsets have an element $a$ where $x >> a$ for all $a \ne a$.

The last is true for $>$ but not for $<$ on the Naturals. The last is true for $<$ but not $>$ on the negative Naturals. The last is not true for either $<$ or $>$ on the integers.