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I am given the following definition:

For an arbitrary order type $\Theta$, denote by $\Theta$* (the reverse of $\Theta$) the order type $type\Theta$*$=typeA(\succ)$, where $\langle A,\prec \rangle$ is an ordered set of order type $\Theta$.

And the example:

$\omega_0$*$=type${$-n : n \in \omega $}$(<)$.

I am having a hard time understanding this. Is the example showing that if $\omega_0$ is the order type of $\omega$, then $\omega_0$* is the order type of the negative integers?

Also, if $\eta_0$ is the order type of $\mathbb{Q}$, why does $\eta_0 = \eta_0$*?

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2 Answers 2

up vote 4 down vote accepted

Yes, $\omega_0^*$ is indeed the order type of the negative integers. The idea is just to "write it upside down" and get a new order type.

Sometimes, however, when writing the order type in reverse we get the same result, for instance if you take $\{0,1,2,3\}$ the natural order $0<1<2<3$ and its inverse $3>2>1>0$ are of the same order type.

Furthermore, if you take the integers you have the same result, that is $\mathbb Z$ with the usual ordering $<$ and with the inverse ordering $>$ are of the same order type (the function $x\mapsto -x$ is the isomorphism needed to show that).

From the above example, one can infer the result for the rational numbers. The inverse order means that $q_1<q_2$ then $q_2>q_1$, and as before $x\mapsto -x$ is the needed isomorphism to prove that.

Added: We say that $a<b$ if $\langle a,b\rangle\in <$, inverse orders are essentially $\langle b,a\rangle$ when $\langle a,b\rangle\in <$.

Suppose $\langle A,R\rangle$ and $\langle B,S\rangle$ are two ordered sets, an order isomorphism $f$ is a function with the following properties:

  1. $f\colon A\to B$ is a bijection;
  2. $aRb\iff f(a)Sf(b)$, that is to say: $\langle a,b\rangle\in R\iff \langle f(a),f(b)\rangle\in S$.

From this follows that an isomorphism preserves properties such as "$a$ is minimal if and only if $f(a)$ is minimal".

This means that $\omega$ and $\omega^*$ are not isomorphic since $0$ is minimal in the non-negative integers, but $f(0)$ is maximal in the non-positive integers.

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How is it that the order type of the positive integers is different than that of the negative integers? What makes it so? What is the defining characteristic of an order type? –  furs Jul 4 '11 at 18:25
    
@furs: In the negative integers not every subset has a minimal element, but every subset has a maximal element. In the positive integers this is exactly the other way around. –  Asaf Karagila Jul 4 '11 at 18:28
    
Is this why a well ordered set can't have subsets of order type $\omega$*? If it did, then this subset wouldn't have a minimal element. Is this correct? –  furs Jul 5 '11 at 5:53
    
@furs: Exactly. –  Asaf Karagila Jul 5 '11 at 6:50
1  
@furs: You will have to require linearity as well; however being well-founded means no infinitely decreasing chain, i.e. no embedding of $\omega^*$. (A well ordering is a well-founded linear order) –  Asaf Karagila Jul 5 '11 at 19:03

Yes, $\omega_0^*$ is the order type of the negative integers, and that's what the example is showing; reversing an order type simply 'turns it around'. To show that $\eta_0^* = \eta_0$, you just have to show that $\langle \mathbb Q, < \rangle$ is order-isomorphic to $\langle \mathbb Q, > \rangle$. Informally, you have to show that 'turning $\mathbb Q$ around' doesn't change the 'shape' of its order.

More precisely, you have to find a bijection $h$ from $\mathbb Q$ to itself such that for all $p,q \in \mathbb Q$, $h(p) > h(q)$ iff $p<q$. You know that $p<q$ iff $-p > -q$, so the function $h:\mathbb Q \to \mathbb Q:p \mapsto -p$ looks like a very good bet. Clearly $h(p) > h(q)$ iff $-p > -q$ iff $p<q$, and I'll leave it to you to verify that $h$ is a bijection on $\mathbb Q$.

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Just a comment: your argument shows that the order type of any ordered abelian group is self-dual in this sense. –  Pete L. Clark Jul 4 '11 at 9:39

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