Wiki mentions the following for order type on even ordinals.
Consider the set of even ordinals less than ω·2+7, which is:
V = {0, 2, 4, 6, ...; ω, ω+2, ω+4, ...; ω·2, ω·2+2, ω·2+4, ω·2+6}.
Its order type is:
ord(V) = ω·2+4 = {0, 1, 2, 3, ...; ω, ω+1, ω+2, ...; ω·2, ω·2+1, ω·2+2, ω·2+3}.
Because there are 2 separate lists of counting and 4 in sequence at the end.
Unfortunately, I'm at a loss understanding why ord(V) = ω·2+4. ω·2+4 appears to be a limit on ord(V), then I at least understand that part and that ord(V) = ω·2+4 = all ranks less then this limit, but that is all.
Does anyone have any insight into what Wiki is trying to show? Also, why does the ordering on the set (total, partial, etc) relate to the infinite limit ordinal ω?
Thanks again!
The order-type of a well-ordered set is the unique ordinal it is order-isomorphic to. In this case, you can read off the order-isomorphism from the way the two sets are listed: $$V = \{0, 2, 4, 6, \dots; \omega, \omega+2, \omega+4, \dots; \omega\cdot2, \omega\cdot2+2, \omega\cdot2+4, \omega\cdot2+6\}$$ $$\omega\cdot2+4 = \{0, 1, 2, 3, \dots; \omega, \omega+1, \omega+2, ...; \omega\cdot2, \omega\cdot2+1, \omega\cdot2+2, \omega\cdot2+3\}.$$ Namely, the isomorphism $f:V\to \omega\cdot 2+4$ is given by $f(0)=0$, $f(2)=1$, $f(4)=2$, $f(6)=3$, and so on, and then $f(\omega)=\omega$, $f(\omega+2)=\omega+1$, $f(\omega+4)=\omega+2$, and so on, and then finally $f(\omega\cdot2)=\omega\cdot2$, $f(\omega\cdot2+2)=\omega\cdot2+1$, $f(\omega\cdot2+4)=\omega\cdot2+2$, and $f(\omega\cdot2+6)=\omega\cdot2+3$. A bit more formally, we can say we define $f(n)=n/2$ and $f(\omega+n)=\omega+n/2$ for each $n<\omega$, and $f(\omega\cdot 2+n)=\omega\cdot 2+n/2$ for $n=0,2,4,6$.
The fact that this isomorphism $f$ exists proves that $ord(V)=\omega\cdot 2+4$, by definition of order-type.