If $\left\langle \alpha_\gamma : \gamma < \beta\right\rangle$ is a sequence of ordinals of length $\beta$, we define the sum $\sum_{\gamma < \beta} \alpha_\gamma$ inductively as follows:
- If $\beta = 0$, $\sum_{\gamma < \beta} \alpha_\gamma := 0$.
- If $\beta = \delta + 1$, $\sum_{\gamma < \beta} \alpha_\gamma := \left(\sum_{\gamma < \delta}\alpha_\gamma\right) + \alpha_\delta$.
- If $\beta$ is a limit ordinal, $\sum_{\gamma < \beta} \alpha_\gamma := \sup_{\delta < \beta} \sum_{\gamma < \delta}\alpha_\gamma$.
Note that the sum depends on the order of the ordinals in the sequence, and I do not require the sequence of ordinals to be increasing.
I would like to know if for all limit ordinals $\alpha$ and ordinal $\beta \geq \mathrm{cf}(\alpha)$, there exists a sequence of ordinals $\left\langle \alpha_\gamma : \gamma < \beta\right\rangle$ such that $\alpha_\gamma < \alpha$ for all $\gamma < \beta$, and $\alpha = \sum_{\gamma < \beta} \alpha_\gamma$?
Best Answer
Recall that for all ordinals $\alpha\leq\beta$ there is a (unique) $\gamma$ with $\alpha+\gamma=\beta$ and this $\gamma$ is $\leq\beta$. So if $\alpha$ is any limit ordinal and $\langle \alpha_i\mid i<\beta\rangle$ is any increasing sequence of ordinals below $\alpha$ converging to $\alpha$ then there is a sequence $\langle \gamma_i\mid i<\beta\rangle$ below $\alpha$ with $$(\mathrm{sup}_{j<i}\alpha_j)+\gamma_i=\alpha_i$$ for all $i<\beta$. An induction shows that for any $i<\beta$ $$\sum_{j<i}\gamma_j=\mathrm{sup}_{j<i}\alpha_j$$ so in particular $$\sum_{i<\beta}\gamma_j=\alpha$$ Thus there is such a sum of smaller ordinals of length $\beta=\mathrm{cof}(\alpha)$. Any longer sum can be achieved by padding with $0$'s.