Order type of a set that is ordered but not well-ordered

Question

I looking at ordered sets in the Kolmogorov (Introductory Read Analysis) book.

Section 3.5 Definition 2 says

the order type of a well ordered set is called an ordinal number. If the set is infinite, the ordinal is transfinite

So the set of positive integers $\{1,2,...,k,..\}$ arranged in increasing order is well ordered with the order type $\omega$.

The set $\{..., -k,.., -2, -1\}$ is ordered but not well-ordered. But this set is isomorphic to the set of positive integers - map each element to its negative.

So if order type is something shared by all isomorphic sets (Section 3.3), why can the set of negative integers (a set that is ordered but not well ordered and is isomorphic to a well ordered set) not have an order type?

More broadly, if a set is ordered but not well ordered, but is isomorphic to a well ordered set, why should it not have an order type?

Answer 1

As already pointed out in the comments, there are some problems with the question details, involving misunderstanding of a bijection as an isomorphism.

The crux of the question was that if a well-ordered set $A$ has an order type, and the set $A$ has a bijection with another set $B$, then $B$ should also have an order type.

This is a somewhat similar confusion as raised in this question.

The answer is in the Mathworld article on this topic:

In general, if $\alpha$ is any order type, then $^*\alpha$ is the same type ordered backwards (Dauben 1990, p. 153).

The order type of the set $Z_-$ is $^*\omega$ or $\omega^*$.

Bijectivity implies the same cardinality. Not the same order type. $Z_-$ is bijective to $Z_+$, so they are equipotent or equipollent, so their cardinalities are the same.

Tarski (1924) proposed to instead define a cardinal number by stating that every set A is associated with a cardinal number |A|, and two sets A and B have the same cardinal number iff they are equipollent

Answer 2

Suppose $ (\Bbb Z_+,\le)$ and $(\Bbb Z_-,\le)$ are order isomorphic.

Then $\exists f:\Bbb{Z}_+ \to \Bbb{Z}_-$ such that

1. $f$ is binjective.

2. $f$ preserve order i.e $a\le b$ implies $f(a) \le f(b) $

Since $|\Bbb{Z}_+|=\aleph_0=| \Bbb{Z}_-|$ , hence there is a bijection between $\Bbb{Z}_+ $ and $\Bbb{Z}_-$.

Claim: No bijection $f: \Bbb{Z}_+ \to \Bbb{Z}_-$ is not order preserving.

Proof: Assume the contrary.

Let $f(1) =u$

Then $\forall x\in \Bbb{Z}_+ \space , 1\le x$ implies $u=f(1) \le f(x) $

i.e $u\le y\space , \forall y\in \Bbb{Z}_-$ implies $u$ is the least element which is a contradiction because $\Bbb{Z}_-$ in it's usual ordering has no least element (given any $u\in \Bbb{Z}_- $ , $u-1<u$ ).

Note: Order isomorphism between two poset maps least element to least element.

Conclusion: $(\Bbb{Z}_+, \le) $ and $( \Bbb{Z}_-, \le) $ is not order isomorphic.