For the first transfinite ordinal $\omega$, we know that it is additively indecomposable, which makes sense since the sum of any two finite ordinals/numbers is, of course, again finite. However, I was wondering about more general sums of the form $$\sum_{i<\alpha}\beta_i=\omega$$ where $\alpha$ and $\beta_i$ (for all $i$) are again ordinals. My question might be rather basic, but I cannot immediately find a source explaining it. Can we find $\alpha$ and $\beta_i$ such that $\omega$ allows a decomposition as above (obviously I do not consider the trivial case $\alpha=1$ and $\beta_0=\omega$)?
Aside from the result in the question Infinite Ordinal Sum, I guess/expect that this also works for $\alpha=\omega$ and $\beta_i$ any finite ordinal for all $i<\omega$?
The reason why I am doubting my idea comes from the fact that $\omega+1\neq\omega$ since the left-hand side has a maximal element (still valid if we replace $1$ by another finite ordinal) and, hence, taking sums with transfinite objects is not as simple as in ordinary arithmetic. However, when we take an ''infinite sum'' of finite ordinals there is not really a ''last'' ordinal, so there probably is also no maximal element...
$\sum_{i\lt\alpha}\beta_i$ will be well-defined (see the first answer in the question you linked to). I believe the following would be true (with a proof long and tedious enough):
[1] Provided all $\beta_i\ne0$, this sum will be $\omega$ if and only if:
[2] If not all $\beta_i\ne 0$, the set $\kappa:=\{i\in\alpha\mid\beta_i\ne0\}$ (the "support" of the map $i\mapsto\beta_i$) is well-ordered, hence an ordinal (up to a bijection preserving ordering), and we can apply the criterion [1] to $\sum_{i\in\kappa}\beta_i$, which is of the same form as the original sum (after a suitable re-arrangement of the indices).
Thanks to @AndrésE.Caicedo for correction of a previous version of this answer (in the comments below).