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

score 167 · Answer 15 · 2018-11-08 01:20:07

A question about the proof of $\sup\{\alpha^\beta\alpha^\delta\mid\delta<\gamma\}=\alpha^\beta\sup\{\alpha^\delta\mid\delta<\gamma\}$

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A question about the proof of $\sup\{\alpha^\beta\alpha^\delta\mid\delta<\gamma\}=\alpha^\beta\sup\{\alpha^\delta\mid\delta<\gamma\}$

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