Order type and its reverse 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 $\eta_0 = \eta_0$*? 
 A: 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:


*

*$f\colon A\to B$ is a bijection;

*$aRb\iff f(a)Sf(b)$, 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.
A: 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 $\mathbb Q$.
