Question regarding addition of order types I have been reading through Herbert Endertons introductory book on set theory as I have stumbled upon a claim that baffled me through the day.
As anyone I need sleep so I ask here for help.
Namely as I was reading about order types the author makes a claim that $\bar1 + \bar\omega = \bar\omega$ but that $\bar\omega + \bar1 = \bar\omega^+$
Where $\bar1$ stands for order type for $\langle1,\epsilon_1\rangle$ and $\bar\omega$ stands for order type of $\langle\omega,\epsilon_\omega\rangle$
I have some how conviced myself that first claim is true but for second one I can not even start to understand it.Can someone give a proof of the claim,or even better an intuitive answer?
Along with the answer for this question an example of addition of some arbitrary order types(along with through description of steps) would 
 A: Ok so here goes the answer that I came up with.
Let $\bar1 = it\langle1,\epsilon_1\rangle$ and $\bar\omega=it\langle\omega,\epsilon_\omega\rangle$
As by definition of addition of order types $\bar1 + \bar\omega$ we require $\langle A,R\rangle\in\bar\omega$ and $\langle B,S \rangle\in\bar1$ where A and B are disjoint sets.
We now define :
$$R\oplus S = R\cup S \cup (A\times B)$$
It is easily provable that defined set is linear ordering on $A\cup B$
Now notice that since B is equinumerous to 1 it must contain only one element,which is maximum in $A \cup B$ by the $R \oplus S$ ordering
Now let $\alpha$ be the single element in B we can now see that $\langle \omega ,\epsilon_\omega \rangle \cong \langle seg\;\alpha,R\rangle$ thus it can not be isomorphic to $\langle A \cup B , R \oplus S\rangle $ thus to find isomorphism we consider the next largest ordinal namely succesor of omega and thus we see it satisfies since it contains and aditional element to pair with our element $\alpha$
Piece of cake right?
