Prove $\sup\limits_{\delta<\gamma}(\alpha+(\beta+\delta))=\sup\limits_{\epsilon<\beta+\gamma}(\alpha+\epsilon)$

ordinals

I'm trying to prove this theorem but stuck at proving $\color{blue}{\epsilon<\beta+\delta'\text{ for some }\delta'<\gamma\implies\epsilon=\beta+\delta\text{ for some }\delta<\gamma}$. Any help is greatly appreciated!

Let $\alpha,\beta,\gamma,\delta,\epsilon$ be ordinals where $\gamma\neq\emptyset$ is limit. Then $$\sup\limits_{\delta<\gamma}(\alpha+(\beta+\delta))=\sup\limits_{\epsilon<\beta+\gamma}(\alpha+\epsilon)$$

My attempt:

Lemma: $\alpha+\beta<\alpha+\gamma\iff \beta<\gamma$ for all ordinals $\alpha,\beta,\gamma$.

Let $A:=\{\alpha+(\beta+\delta)\mid\delta<\gamma\}=\{\alpha+(\beta+\delta)\mid\beta+\delta<\beta+\gamma\}$ by Lemma, and $B:=\{\alpha+\epsilon\mid\epsilon<\beta+\gamma\}$. Our task is to prove $\sup A=\sup B$.

It's clear that $A\subseteq B$, so our task is done if we show $B\subseteq A$.

For $b\in B$, $b=\alpha+\epsilon$ for some $\epsilon<\beta+\gamma$. Since $\gamma$ is limit, $\beta+\gamma$ is limit. Thus $\beta+\gamma=\sup\limits_{\delta<\gamma}(\beta+\delta)$. Then $b=\alpha+\epsilon$ for some $\epsilon<\sup\limits_{\delta<\gamma}(\beta+\delta)$.

We have $\epsilon<\sup\limits_{\delta<\gamma}(\beta+\delta)\implies \color{blue}{\epsilon<\beta+\delta'\text{ for some }\delta'<\gamma\implies\epsilon=\beta+\delta\text{ for some }\delta<\gamma}$.

Best Answer

I have found a way to fix my previous issue and posted here.

Lemma 1: Let $A$ be the set of ordinals, $B\subseteq A$ such that $\forall\alpha\in A,\exists\beta\in B:\alpha\le\beta$. Then $\sup A=\sup B$.

Proof:

Show $\{\gamma \mid \forall\alpha\in A : \gamma \ge\alpha\} = \{\gamma \mid \forall \beta\in B: \gamma \ge\beta\}$ and thus the minima of these sets, and hence the suprema of $A$ and $B$, are equal.

Suppose $\forall\alpha\in A : \gamma \ge\alpha$. Then, since $B\subseteq A, \forall \beta\in B: \gamma \ge\beta$.

Suppose $\forall \beta\in B: \gamma \ge\beta$. For any $\alpha\in A$, there is some $\beta\in B$ such that $\alpha\le\beta$, hence $\gamma \ge\beta\ge\alpha$. Thus $\gamma\ge\alpha$ for all $\alpha\in A$.

Lemma 2: $\alpha+\beta<\alpha+\gamma\iff \beta<\gamma$ for all ordinals $\alpha,\beta,\gamma$.

We proceed to prove our theorem.

Let $B:=\{\alpha+(\beta+\delta)\mid\delta<\gamma\}=\{\alpha+(\beta+\delta)\mid\beta+\delta<\beta+\gamma\}$ by Lemma 2, and $A:=\{\alpha+\epsilon\mid\epsilon<\beta+\gamma\}$. Our task is to prove $\sup A=\sup B$.

It's clear that $B\subseteq A$. By Lemma 1, our task is done if we show $\forall a\in A, \exists b\in B:a\le b$.

Let $\alpha+\epsilon\in A$ and $\delta:=\min\{\xi\mid\beta+\xi\ge\epsilon\}$.

  • $\delta\le\gamma$. If not, $\delta>\gamma$ and thus $\gamma>\beta+\epsilon$, which is a contradiction.

  • $\delta<\gamma$. If not, $\delta=\gamma$. For each $\xi<\gamma,\xi<\delta$ and thus $\beta+\xi<\epsilon$. Then $\epsilon<\beta+\gamma=\sup\limits_{\xi<\gamma}(\beta+\xi)\le\sup\limits_{\xi<\gamma}(\epsilon)=\epsilon$, which is a contradiction.

  • By construction, $\beta+\delta\ge\epsilon$ and thus $\alpha+\epsilon\le\alpha+(\beta+\delta)\in B$ by Lemma 2. Thus the conditions of Lemma 1 are satisfied.

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