On Page 44, Set theory, Jech(2006)
(Show that) there are at least $\mathfrak{c}$ countable order-types of linearly ordered sets. [For every sequence $a = \{ a_n : n \in N\}$ of natural numbers consider the order-type $\tau_a=a_0+\xi+a_1+\xi+a_2+\dots$, where $\xi$ is the order type of the integers. Show that if $a \ne b$, then $\tau_a \ne \tau_b$. ($\mathfrak{c}$ is the cardinality of $R$) ]
On Page 19, two sets have the same order type, if and only if they are isomorphic in the sense that there is a bijective function$f$, and both $f$ and $f^{-1}$ are order-preserving. So I speculate that the answer should equal the cardinality of the set of initial segments of the least uncountable ordinal, which is $\omega_1$. I must have misunderstood the problem somehow.
Even more confusing to me is the hint. Allow me to follow the convention in Jech's textbook:
- $1+\omega=\omega \ne \omega+1$
- $2·\omega=\omega \ne \omega·2 =\omega+\omega$
Since the order-type of integers is the sum of that of negative integers and that of natural number, which is $\omega·2$ . Then we have: $\tau_a = (a_0+\omega)+\omega+(a_1+\omega)+……=\omega^2$, which is a constant function, just contrary to injective function as shall be shown.