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Let $\alpha$ be an ordinal and let $\kappa$ be a limit ordinal. I want to prove that $$\alpha+\kappa=\bigcup_{\gamma\in\kappa}\left(\alpha+\gamma\right).$$ Here I define the sum of ordinals not using recursion. I define the sum of two ordinals $\alpha$ and $\beta$ as follows. Choose well-ordered sets $A$ and $B$ which are isomorphic to $\alpha$ and $\beta$, respectively, take their union $A\cup B$ and define a well-order on it by making every element of $A$ less than every element of $B$. Then the unique ordinal which is isomorphic to $A\cup B$ is $\alpha+\beta$. I want to use this direct definition to prove the question, no other results. I am having trouble with it because it is taking the union of well ordered sets makes it very complicated.

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The $\bigcup$ is nested union only adding elements that are larger than everything we have before, so there is nothing complicated about this union.

Both LHS and RHS are partial order on $\alpha\amalg\kappa$, and it is clear that they are well-ordering on the subset of elements comparable with $\alpha$.

LHS is declaring everything in the $\alpha$ copy is less than in $\kappa$ copy.

RHS is declaring everything in the $\alpha$ copy is less than everything inside an element in the $\kappa$ copy (which is what the $\gamma\in\kappa$ is doing). But $\kappa$ being a limit ordinal means $\gamma\in\kappa\Rightarrow\gamma\in\gamma+1\in\kappa$, and so we have each $\gamma\in\kappa$ is also declared larger than everything in $\alpha$. So LHS and RHS defines the same ordering on $\alpha\amalg\kappa$.

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For each proper initial segment $C$ of $B$ we have a unique order isomorphism between $C$ and an ordinal $\gamma \in \kappa$.

This gives us a unique order isomorphism $f_C : A \cup C \to \alpha + \gamma$, and by uniqueness all of these isomorphisms agree on their common domain.

So we can take the union $f = \bigcup f_C$, which is an order isomorphism between $A \cup B$ and $\bigcup_{\gamma \in \kappa}(\alpha + \gamma)$.

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