Let $\alpha,\gamma$ be ordinals and $A$ be a set of ordinals. Then $\gamma<\alpha+\sup_{\beta\in A}\beta\implies\exists\beta\in A:\gamma<\alpha+\beta$

Question

Let $\alpha,\gamma$ be ordinals and $A$ be a set of ordinals. Then $$\gamma<\alpha+\sup\limits_{\beta\in A}\beta\implies\exists\beta\in A:\gamma<\alpha+\beta$$

This problem arises when I prove the theorem:

Let $\alpha,\beta$ be ordinals and $A$ be a set of ordinal. Then $\sup\limits_{\beta\in A}(\alpha+\beta)=\alpha+\sup\limits_{\beta\in A}(\beta)$

I'm almost done except for one minor point: $\gamma<\alpha+\sup_\limits{\beta\in A}\beta\implies\exists\beta\in A:\gamma<\alpha+\beta$. This result seems quite obvious, but I don't know how to prove in spite of several attempts.

Could you please shed me some lights on how to prove $\gamma<\alpha+\sup\limits_{\beta\in A}\beta\implies\exists\beta\in A:\gamma<\alpha+\beta$? Thank you very much!

Answer 1

Let $\sigma=\sup\limits_{\beta\in A}\beta$.

• If $\sigma\in A$, then $\sigma=\beta$ for some $\beta\in A$. Then $\gamma<\alpha+\sup_\limits{\beta\in A}\beta=\alpha+\sigma=\alpha+\beta$ for some $\beta\in A$.

• If $\sigma\notin A$, then clearly $\sigma$ is a limit ordinal and thus $\sigma=\sup_\limits{\eta<\sigma}\eta$. Then $\gamma<\alpha+\sup_\limits{\beta\in A}\beta=\alpha+\sigma=$ $\alpha+\sup_\limits{\eta<\sigma}\eta=\sup_\limits{\eta<\sigma}(\alpha+\eta)$. It follows that $\gamma<\sup_\limits{\eta<\sigma}(\alpha+\eta)$ and thus $\gamma<\alpha+\eta$ for some $\eta<\sigma$. Moreover, $\eta<\sigma\implies\eta < \sup\limits_{\beta\in A}\beta\implies\eta<\beta$ for some $\beta\in A$. Thus $\alpha+\eta<\alpha+\beta$ for some $\beta\in A$ and hence $\gamma<\alpha+\eta<\alpha+\beta$ for some $\beta\in A$. As a result, $\exists\beta\in A:\gamma<\alpha+\beta$.

Answer 2

Let $α$ be an ordinal and $A$ be a set of ordinals. Then $\sup\limits_{β∈A}(α\cdot β) = α\cdot\sup\limits_{β∈A}(β)$.

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My attempt:

Let $X=\{\gamma\mid\gamma<\sup\limits_{β∈A}(α\cdot β)\}$, $Y=\{\gamma\mid\gamma<α\cdot\sup\limits_{β∈A}(β)\}$, and $\sigma=\sup\limits_{β∈A}(β)$.

For $\gamma\in X:\gamma<\sup\limits_{β∈A}(α\cdot β)$ and thus $\gamma<α\cdot β$ for some $\beta\in A$. It follows that $\gamma<α\cdot\sup\limits_{β∈A}(β)$ and thus $\gamma\in Y$. Hence $X\subseteq Y$.

For $\gamma\in Y:\gamma<α\cdot\sup\limits_{β∈A}(β)=\alpha\cdot\sigma$.

• If $\sigma\in A$, then $\sigma=\beta$ for some $\beta\in A$. It follows that $\gamma<\alpha\cdot\sigma=\alpha\cdot\beta\le\sup\limits_{β∈A}(α\cdot β)$ for some $\beta\in A$. Hence $\gamma<\sup\limits_{β∈A}(α\cdot β)$ and thus $\gamma\in X$. As a result, $Y\subseteq X$.

• If $\sigma\notin A$, then $\sigma$ is clearly a limit ordinal and thus $\sigma=\sup\limits_{\eta<\sigma}\eta$. It follows that $\gamma<\alpha\cdot\sigma=\alpha\cdot\sup\limits_{\eta<\sigma}\eta=$ $\sup\limits_{\eta<\sigma}(\alpha\cdot\eta)$. Thus $\gamma<\sup\limits_{\eta<\sigma}(\alpha\cdot\eta)$ and hence $\gamma<\alpha\cdot\eta$ for some $\eta<\sigma$. Since $\eta<\sigma=\sup\limits_{β∈A}(β)$, $\eta<\beta$ for some $\beta\in A$ and thus $\alpha\cdot\eta<\alpha\cdot\beta$ for some $\beta\in A$. As a result, $\gamma<\alpha\cdot\eta<\alpha\cdot\beta\le\sup\limits_{β∈A}(α\cdot β)$ for some $\beta\in A$. Hence $\gamma<\sup\limits_{β∈A}(α\cdot β)$ and thus $\gamma\in X$. As a result, $Y\subseteq X$.

To sum up, $X=Y$ and thus $\sup\limits_{β∈A}(α\cdot β) = α\cdot\sup\limits_{β∈A}(β)$.