TITLE: Trying to understand the sum of ordinals
QUESTION [5 upvotes]: I'm having a hard time with the functions defined by transfinite induction, in particular I have the sum of two ordinals defined as follows:
\begin{align}\alpha+0&=\alpha \\ 
\alpha+s(\beta)&=s(\alpha+\beta)\\ 
\alpha+\beta&=\sup_{\gamma<\beta}(\alpha+\gamma)\quad \mbox{(if $\beta$ is a limit ordinal)} \end{align}
I've noticed that if $\alpha<\omega$ then $\alpha + \omega =\omega$ and it seems to me that if I replace $\omega$ by any other countable limit ordinal that assertion still holds, is this right? And, can it be generalized to any limit ordinal?
I feel I'm still not able to grasp the intuition behind it.

REPLY [7 votes]: The intuition is actually quite a bit simpler than the recursive definition. For any linear orders $\langle A,\le_A\rangle$ and $\langle B,\le_B\rangle$, not necessarily well orders, we can define a linear order that is their sum: the underlying set is $(\{0\}\times A)\cup(\{1\}\times B)$, and the order $\le$ is lexicographic: $\langle i,x\rangle\le\langle j,y\rangle$ iff either $i<j$, or $i=j=0$ and $x\le_A y$, or $i=j=1$ and $x\le_B y$. Essentially we’re just putting $B$ with its given linear order after $A$ with its linear order:
$$\overset{A}\longrightarrow\overset{B}\Longrightarrow\;.$$
The bit with the $0$ and $1$ is a technical detail to ensure that we’re working with disjoint sets; it’s unnecessary if $A$ and $B$ are disjoint.
In the case of ordinals $\alpha$ and $\beta$, we can form a sum in just this fashion: the underlying set is $(\{0\}\times\alpha)\cup(\{1\}\times\beta)$, and we have $\langle i,\xi\rangle\le\langle j,\eta\rangle$ iff $i<j$, or $i=j$ and $\xi\le \eta$. It’s not at all hard to check that this order $\le$ is a well-order, so it is order-isomorphic to a unique ordinal $\gamma$, and we write $\alpha+\beta=\gamma$. Intuitively, $\gamma$ is the ordinal whose order type is obtained by sticking a copy of $\beta$ after a copy of $\alpha$.
It might be worthwhile to try proving to yourself that this more intuitive approach yields the same result as the recursive definition.
When $\alpha$ is finite and $\beta$ is not, $\alpha$ together with the first $\omega$ elements of $\beta$ still form just one copy of $\omega$, so — as you said — in this case $\alpha+\beta=\beta$. This can be generalized, but it takes a bit of care. Let $\alpha_0=\alpha$. Given $\alpha_n$ for some $n\in\omega$, let $\alpha_{n+1}=\alpha_n+\alpha$. Finally, let $\eta=\sup_n\alpha_n$. (In fact $\eta=\alpha\cdot\omega$, which the multiplication is ordinal multiplication.) Then $\alpha+\beta=\beta$ whenever $\beta\ge\eta$, because one more $\alpha$ just gets absorbed into the initial $\eta$ part of $\beta$.