I'm reading a somewhat old fashionned book on Set Theory ( first edition 1964).
I think I have understood the concept of ordinal number : if A is well ordered, then the order type of A ( its equivalence class under the relation of similarity) is A's cardinal number.
I'm now trying to understand operations on ordinals, and firstly, addition.
My author says that: if A and B are disjoint and if a= ord(A) and b=ord(B), then
a+b = ord ( {A,B} ).
With some aid from another book, namely Pinter's Book of Set Theory ( and from answers on this site), I understood that
{A,B} is, so to say, the ordered union of A and of B under the relation R such that xRy ( or : (x,y) belongs to R if you prefer) :
(1) x and y belong to A and x precedes ( or is equal to) y
(2) x and y belong to B and x precedes ( or is equal to) y
(3) x belongs to A and y belongs to B.
At this point my author gives an example aiming at illustrating addition of ordinals ( please, see image below).
My question : why does he say that in one case " n+omega = omega" and that , in the other case " omega + n is greater that omega".
Remark : my question is not on non-commutativity of ordinal addition; I am perfectly ready to accept this fact; what I do not understand is the reason why the "results" are not the same in the two cases
Remark : the author uses the symbols P and " omega" defined as follows :
P : the set of counting numbers
"omega" : ordinal number of P ( the set of counting numbers)
It is not true that if $A$ is well ordered then the order type of $A$ is its cardinal number. The well ordered set $\{1,2,3,4,5,\ldots,z\}$ has order type $\omega+1$, but its cardinal is $\aleph_0$ (or $\omega$ if we use it as a cardinal). It is an example of the addition you are talking about. We started with $\omega$ and put one element after it. If we put the new element before it we would have $\{z,1,2,3,4,5,\ldots\}$ which has the same order type as $\omega$. This is $1+\omega$ because we put the single element first, and shows and example of $1+\omega=\omega$