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

Question

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

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

Answer 1

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

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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.

Answer 2

I have presented proofs for some related results.

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Theorem 1: Let $\alpha,\beta,\gamma,\delta,\epsilon$ be ordinals and $\gamma$ is limit. Then $\sup\limits_{\delta<\beta+\gamma}(\alpha\cdot\delta)=\sup\limits_{\beta+\epsilon<\beta+\gamma}(\alpha\cdot(\beta+\epsilon))$.

Theorem 2: Let $\alpha,\beta,\gamma,\delta,\epsilon$ be ordinals and $\gamma$ is limit. Then $\sup\limits_{\alpha\cdot\epsilon<\alpha\cdot\gamma}(\alpha\cdot\beta+\alpha\cdot\epsilon)=\sup\limits_{\zeta<\alpha\cdot\gamma}(\alpha\cdot\beta+\zeta)$.

Theorem 3: Let $\alpha,\beta,\gamma,\delta,\epsilon$ be ordinals and $\gamma$ is limit. Then $\sup\limits_{\beta\cdot\epsilon<\beta\cdot\gamma}(\alpha\cdot(\beta\cdot\epsilon))=\sup\limits_{\delta<\beta\cdot\gamma}(\alpha\cdot\delta)$.

Theorem 4: Let $\alpha,\beta,\gamma,\delta,\epsilon$ be ordinals and $\gamma$ is limit. Then $\sup\limits_{\delta<\beta+\gamma}(\alpha^\delta)=\sup\limits_{\beta+\epsilon<\beta+\gamma}(\alpha^{\beta+\epsilon})$.

Theorem 5: Let $\alpha,\beta,\gamma,\delta,\epsilon$ be ordinals and $\gamma$ is limit. Then $\sup\limits_{\alpha^\epsilon<\alpha^\gamma}(\alpha^\beta\cdot\alpha^\epsilon)=\sup\limits_{\zeta<\alpha^\gamma}(\alpha^\beta\cdot\zeta)$.

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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 $\alpha,\beta,\gamma$.

Lemma 3: $\alpha\cdot\beta<\alpha\cdot\gamma\iff \beta<\gamma\wedge\alpha\neq 0$ for all ordinals $\alpha,\beta,\gamma$.

Lemma 4: $\alpha^\beta<\alpha^\gamma\iff \beta<\gamma\wedge\alpha>1$ for all ordinals $\alpha,\beta,\gamma$.

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1. Theorem 1.

Let $B:=\{\alpha\cdot(\beta+\epsilon)\mid\beta+\epsilon<\beta+\gamma\}$ and $A:=\{\alpha\cdot\delta\mid\delta<\beta+\gamma\}$.

Clearly, $B\subseteq A$. Let $\alpha\cdot\gamma\in A$ and $\epsilon=\min\{\eta\mid\beta+\eta\ge\delta\}$.

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

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

• $\beta+\epsilon<\beta+\gamma$ by Lemma 2.

• By construction, $\delta\le\beta+\epsilon$ and thus $\alpha\cdot\delta\le\alpha\cdot(\beta+\epsilon)$ by Lemma 3. Thus the conditions of Lemma 1 are satisfied.

2. Theorem 2.

Let $B:=\{\alpha\cdot\beta+\alpha\cdot\epsilon\mid\alpha\cdot\epsilon<\alpha\cdot\gamma\}$ and $A:=\{\alpha\cdot\beta+\zeta\mid\zeta<\alpha\cdot\gamma\}$.

Clearly, $B\subseteq A$. Let $\alpha\cdot\beta+\zeta\in A$ and $\epsilon=\min\{\eta\mid\alpha\cdot\eta\ge\zeta\}$.

• $\epsilon\le\gamma$. If not, $\epsilon>\gamma$ and thus $\alpha\cdot\gamma<\zeta$, which is a contradiction.

• $\epsilon<\gamma$. If not, $\epsilon=\gamma$. Then for each $\eta<\gamma,\eta<\epsilon$ and thus $\alpha\cdot\eta<\zeta$. Then $\zeta<\alpha\cdot\gamma=\sup\limits_{\eta<\gamma}(\alpha\cdot\eta)\le\sup\limits_{\eta<\gamma}(\zeta)=\zeta$, which is a contradiction.

• $\alpha\cdot\epsilon<\alpha\cdot\gamma$ by Lemma 3.

• By construction, $\alpha\cdot\epsilon\ge\zeta$ and thus $B\ni\alpha\cdot\beta+\alpha\cdot\epsilon\ge\alpha\cdot\beta+\zeta$ by Lemma 2. Thus the conditions of Lemma 1 are satisfied.

3. Theorem 3.

Let $B:=\{\alpha\cdot(\beta\cdot\epsilon)\mid\beta\cdot\epsilon<\beta\cdot\gamma\}$ and $A:=\{\alpha\cdot\delta\mid\delta<\beta\cdot\gamma\}$.

Clearly, $B\subseteq A$. If $A=\emptyset, B=A$.

Let $\alpha\cdot\delta\in A$ and $\epsilon=\min\{\zeta\mid\beta\cdot\zeta\ge\delta\}$.

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

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

• $\beta\cdot\epsilon<\beta\cdot\gamma$ by Lemma 3.

• By construction, $\beta\cdot\epsilon\ge\delta$ and thus $B\ni\alpha\cdot(\beta\cdot\epsilon)\ge\alpha\cdot\delta$ by Lemma 3. Thus the conditions of Lemma 1 are satisfied.

4. Theorem 4.

Let $B:=\{\alpha^{\beta+\epsilon}\mid\beta+\epsilon<\beta+\gamma\}$ and $A:=\{\alpha^\delta\mid\delta<\beta+\gamma\}$.

Clearly, $B\subseteq A$. Let $\alpha^\delta\in A$ and $\epsilon=\min\{\eta\mid\beta+\eta\ge\delta\}$.

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

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

• $\beta+\epsilon<\beta+\gamma$ by Lemma 2.

• By construction, $\beta+\epsilon\ge\delta$ and thus $B\ni\alpha^{\beta+\epsilon}\ge\alpha^\delta$ by Lemma 4. Thus the conditions of Lemma 1 are satisfied.

5. Theorem 5.

Let $B:=\{\alpha^\beta\cdot\alpha^\epsilon\mid\alpha^\epsilon<\alpha^\gamma\}$ and $A:=\{\alpha^\beta\cdot\zeta\mid\zeta<\alpha^\gamma\}$.

Clearly, $B\subseteq A$. Let $\alpha^\beta\cdot\zeta\in A$ and $\epsilon=\min\{\eta\mid\alpha^\eta\ge\zeta\}$.

• $\epsilon\le\gamma$. If not, $\epsilon>\gamma$ and thus $\alpha^\gamma<\zeta$, which is a contradiction.

• $\epsilon<\gamma$. If not, $\epsilon=\gamma$. Then for each $\eta<\gamma,\eta<\epsilon$ and thus $\alpha^\eta<\zeta$. Then $\zeta<\alpha^\gamma=\sup\limits_{\eta<\gamma}(\alpha^\eta)\le\sup\limits_{\eta<\gamma}(\zeta)=\zeta$, which is a contradiction.

• $\alpha^\epsilon<\alpha^\gamma$ by Lemma 4.

• By construction, $\alpha^\epsilon\ge\zeta$ and thus $\alpha^\beta\cdot\alpha^\epsilon\ge\alpha^\beta\cdot\zeta$ by Lemma 3. Thus the conditions of Lemma 1 are satisfied.

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L1: If $α$ and $β$ are ordinals, and $α ≤ β$, then $α +1 ≤ β +1$.

L2: If $α$ and $β$ are ordinals, then $α ≤ α + β$.

L3: If $α$ and $β$ are ordinals, then $β ≤ α + β$.

L4: If $γ$ is a limit, then for all $α: α + γ$ is a limit.

L5: Suppose $γ$ is a limit, $α,β$ are ordinals and $β < γ$. then $α + β < α + γ$.

L6: Suppose $α,β,γ$ are ordinals and $β < γ$. Then $α + β < α + γ$.

L7: (Left-Monotonicity of Ordinal Addition). Let $α,β,γ$ be ordinals. The following are equivalent:

i. $β < γ$.

ii. $α + β < α + γ$.

L8: If $α$ and $β\neq 0$ are ordinals, then $α ≤ α · β$.

L9: If $α \neq 0$ and $β$ are ordinals, then $β ≤ α · β$.

L10: If $γ$ is a limit, then for all $α \neq 0: α · γ$ is a limit.

L11: Suppose $γ$ is a limit, $α \neq 0$ and $β$ are ordinals and $β < γ$. Then $α · β < α · γ$.

L12: Suppose $α \neq 0$ and $β,γ$ are ordinals and $β < γ$. Then $α · β < α · γ$.

L13: (Left-Monotonicity of Ordinal Multiplication). Let $α,β,γ$ be ordinals. The following are equivalent:

i. $β < γ ∧ α > 0$.

ii. $α · β < α · γ$.

L14: If $α$ and $β \neq 0$ are ordinals, then $α ≤ α^β$.

L15: If $α > 1$ and $β$ are ordinals, then $β ≤ α^β$.

L16: If $γ$ is a limit, then for all $α > 1: α^γ$ is a limit.

L17: Suppose $γ$ is a limit, $α > 1$ and $β$ are ordinals and $β < γ$. Then $α^β < α^γ$.

L18: Suppose $α > 1$ and $β,γ$ are ordinals and $β < γ$. Then $α^β < α^γ$.

L19: (Left-Monotonicity of Ordinal Exponentiation). Let $α,β,γ$ be ordinals and $α > 0$. The following are equivalent:

i. $β < γ ∧ α > 1$.

ii. $α^β < α^γ$.

L20: (Associativity of Ordinal Addition). Let $α,β,γ$ be ordinals. Then $(α + β) + γ = α + (β + γ)$.

L21: (Distributivity). Let $α,β,γ$ be ordinals. Then $α · (β + γ) = α · β + α · γ$.

L22: (Associativity of Ordinal Multiplication). Let $α,β,γ$ be ordinals. Then $(α · β) · γ = α · (β · γ)$.

L23: Let $α,β,γ$ be ordinals. Then $α^{β+γ} = α^β · α^γ$.

L24: Let $α,β,γ$ be ordinals. Then ${\left(α^β\right)}^γ = α^{β · γ}$.