If $\gamma$ is an indecomposable ordinal and $X\subseteq\gamma$, then the order type of $X$ or of its complement must be $\gamma$ Let $\gamma$ be a limit ordinal.
Then $\quad\forall \alpha, \beta<\gamma \left(\alpha+\beta<\gamma \right)\quad$ implies $\quad\forall X\subseteq \gamma \left(\text{type}(X)=\gamma \vee \text{ type}(\gamma \setminus X)=\gamma \right)$.
This is a part of an exercise of Kunen's "Set Theory". I can't prove it. I think I'm missing something trivial.
(The solution is not discussed in this question: Equivalent characterizations of ordinals of the form $\omega^\delta$ )
 A: This is true for every indecomposable ordinal, not just indecomposable limit ordinals. The proof is by induction on the indecomposable ordinal $\gamma.$ I assume you know that the nonzero indecomposable ordinals are just the ordinals of the form $\omega^\alpha.$ The cases $\gamma=0$ and $\gamma=1$ are trivial, so there are only two cases to consider: either $\gamma=\omega^\lambda$ where $\lambda$ is a limit ordinal, or $\gamma=\omega^{\beta+1}$ for some ordinal $\beta.$
Case 1. $\gamma=\omega^\lambda$ where $\lambda$ is a limit ordinal.
Let a partition $\gamma=X_1\cup X_2$ be given. For each $\alpha\lt\lambda,$ the inductive hypothesis applied to the indecomposable ordinal $\omega^\alpha$ implies that $\operatorname{type}(X_i\cap\omega^\alpha)=\omega^\alpha$ for some $i\in\{1,2\};$ in other words, $\lambda=A_1\cup A_2,$ where $A_i=\{\alpha\in\lambda:\operatorname{type}(X_i\cap\omega^\alpha)=\omega^\alpha\}.$ Choose $i\in\{1,2\}$ so that $\sup A_i=\lambda;$ then $\operatorname{type}(X_i)\ge\operatorname{type}(X_i\cap\omega^\alpha)=\omega^\alpha$ for each $\alpha\in A_i,$ so $\operatorname{type}(X_i)\ge\sup\{\omega^\alpha:\alpha\in A_i\}=\omega^\lambda=\gamma.$
Case 2. $\gamma=\omega^{\beta+1}=\omega^\beta\cdot\omega$ for some ordinal $\beta.$
Let a partition $\gamma=X_1\cup X_2$ be given. Write $\gamma=\bigcup_{n\in\omega}E_n$ where $E_n=\left[\omega^\beta\cdot n,\ \omega^\beta\cdot(n+1)\right).$ Since $\operatorname{type}(E_n)=\omega^\beta,$ the inductive hypothesis applied to the indecomposable ordinal $\omega^\beta$ implies that for each $n\in\omega$ we have $\operatorname{type}(X_i\cap E_n)=\omega^\beta$ for some $i\in\{1,2\}.$ Choose $i\in\{1,2\}$ so that $\operatorname{type}(X_i\cap E_n)=\omega^\beta$ for infinitely many $n;$ then $\operatorname{type}(X_i)\ge\omega^\beta\cdot\omega=\gamma.$
Remark. The Hessenberg natural sum $\alpha\oplus\beta$ is the greatest ordinal which can be partitioned into a set of order type $\alpha$ and a set of order type $\beta.$ The statement we have just proved amounts to the implication
$$\forall\alpha,\beta\lt\gamma\ (\alpha+\beta\lt\gamma)\implies\forall\alpha,\beta\lt\gamma\ (\alpha\oplus\beta\lt\gamma).$$
The natural sum of two ordinals can be described explicitly in terms of their Cantor normal forms.
