Ordinal arithmetic help

Question

Let $\alpha$ be an ordinal and $\beta$ a limit ordinal. I wish to prove that $\alpha + \beta$ is a limit ordinal.

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I am using the following definition:

$\gamma$ is a limit ordinal if $\gamma$ is not 0 and is not a successor ordinal.

Ordinal addition:

1. $\alpha + 0 = \alpha$

2. $\alpha +\beta^+ = (\alpha+\beta)^{+}$

3. $\alpha + \lambda = \bigcup\limits_{\beta < \lambda}{(\alpha+\beta)}$

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Here is my attempt:

Since $\beta$ is a limit ordinal, $\beta \neq 0$. Thus, $\alpha + \beta \neq 0$. I just have to now show that $\alpha + \beta$ is not a successor ordinal. Suppose for a contradiction that $\alpha + \beta = \delta^+$ for some ordinal $\delta$. By definition of ordinal addition, $\bigcup\limits_{\lambda < \beta}{(\alpha+\lambda)} = \alpha + \beta =\delta^+$

$\delta \in \bigcup\limits_{\lambda < \beta}{(\alpha+\lambda)}$

$\exists \lambda_0 <\beta$ such that $\delta \in \alpha + \lambda_0$

I am not sure on how to reach a contradiction.

Answer 1

First a tiny fact: if $\beta$ is a limit ordinal, then $\delta< \beta$ implies $\delta^+ < \beta$; if not, then $\delta^+\geq \beta$, and since $\delta^+$ is the smallest ordinal larger than $\delta$, we have $\delta^+=\beta$, a contradiction.

You've found that $\delta \leq \alpha+\lambda_0$ for some $\lambda_0< \beta$. Thus, $\lambda_0^+ <\beta$, and so $$\delta^+ \leq (\alpha+\lambda_0)^+ = \alpha+\lambda_0^+ < \beta$$ giving a contradiction.

Answer 2

Since $\gamma$ is a limit ordinal, it has no largest member. (If $m=\max \gamma$ then $\gamma=m^+$.) So $y\in \gamma\implies y+1\in \gamma.$

If $x\in \alpha +\gamma$ then $x$ is not the largest member of $\alpha +\gamma.$ Because $x\in \alpha +y$ for some $y\in \gamma,$ so $$x<\alpha+y\in\alpha +(y+1)\in \alpha +\gamma.$$ That is, $$x<\alpha +y<\alpha +(y+1)\in \alpha + \gamma.$$

Since $\alpha +\gamma$ has no largest member it cannot be a successor.