Equivalent definition of ordinal sum

by Seth Alire   Last Updated June 12, 2019 08:20 AM

I'm trying to prove that two definitions of ordinal sum are equivalent.

In the present question I want to prove that given $$\tag{1} \alpha+\beta:=\operatorname{ord}(\{0\}\times \alpha\cup \{1\}\times \beta) $$ it follows that $$ \alpha+ \operatorname{S}(\beta)=\operatorname{S}(\alpha+\beta) \text{, where }\operatorname{S}(A):=A\cup\{A\}. $$

Here is my start: \begin{align} \alpha+ \operatorname{S}(\beta)&=\alpha+ (\beta\cup \{\beta\})\\ &=\operatorname{ord}\big(\left(\{0\}\times \alpha\right)\cup \left(\{1\}\times (\beta\cup \{\beta\})\right)\big)\\ &=\operatorname{ord}\big(\left(\{0\}\times \alpha\right)\cup \left(\{1\}\times \beta\right)\cup \left(\{1\}\times \{\beta\}\right)\big)\\ &=\cdots\tag{*}\\ &=\operatorname{ord}\big(\left(\{0\}\times \alpha\right)\cup \left(\{1\}\times \beta\right)\big)\cup \left\{\operatorname{ord}\big(\left(\{0\}\times \alpha\right)\cup \left(\{1\}\times \beta\right)\big)\right\}\\ &=(\alpha+\beta)\cup\{\alpha+\beta\}=\operatorname{S}(\alpha+\beta) \end{align}

I proved the result $$ \operatorname{ord}(A\cup B)=\operatorname{ord} A+\operatorname{ord} B, $$ but $A,B$ have to be woset and disjoint, additionally $A\cup B$ has to be woset.

How do I fill the gap? Are there already error? Perhaps there are easier ways (transfinite induction?), but I would like to know if there is some technical trick to fill (*).


I changed my mind... any hint to any type of proof showing the equivalence of (1) with the following definition (2) is welcomed (incl. ref duplicate / ref. textbook):

  1. $\alpha+0=\alpha$
  2. $\alpha+ \operatorname{S}(\beta)=\operatorname{S}(\alpha+\beta)$
  3. if $\beta$ is a limit ordinal then $\alpha+\beta$ is the limit of the $\alpha+\delta$ for all $\delta<\beta$.

Related Questions

Alternative definition of ordinal - Proof

Updated June 09, 2019 15:20 PM

Continuity of the ordinal addition

Updated June 21, 2019 08:20 AM