Exercise 6.2.7

Proof. To the contrary, suppose that $\alpha +1>\beta$. Then by the definition of $<$ we have that $\beta \in \alpha +1=\alpha \cup \left\{\alpha \right\}$ and since $\beta \ne \alpha$ it has to be that $\beta \in \alpha$. But then $\beta <\alpha$, which is a contradiction. □

Proof. Since $\alpha <\beta +1$ we have that $\alpha \in \beta +1=\beta \cup \left\{\beta \right\}$. Hence $\alpha \in \beta$ or $\alpha =\beta$. Thus $\alpha <\beta$ or $\alpha =\beta$, i.e. $\alpha \le \beta$. □

Proof. Suppose that $X$ is a set of ordinals with no greatest element. Let $\beta =\sup <!--\; FUNCTION\; APPLICATION\; -->X=\bigcup <!--\; FUNCTION\; APPLICATION\; -->X$. Then by the remarks following the proof of Theorem 6.2.6 $\beta \notin X$ since $X$ has no greatest element. Now also suppose that $\beta$ is a successor so that there is an ordinal $\alpha$ such that $\beta =\alpha +1$.

If $\alpha \in X$ then since $X$ has no greatest element there is a $\gamma \in X$ such that $\alpha <\gamma$. Then by Lemma 1 $\beta =\alpha +1\le \gamma$. It cannot be that $\gamma =\beta$ since $\gamma \in X$ but $\beta \notin X$ so it must be that $\beta <\gamma$. But then since $\beta$ is an upper bound of $X$ it follows that $\gamma$ is also. However, since $\gamma \in X$ this would make $\gamma$ the greatest element of $X$, which is a contradiction.

On the other hand if $\alpha \notin X$ then consider any $\gamma \in X$. Then $\gamma <\beta =\alpha +1$ so that by Lemma 2 $\gamma \le \alpha$. Since $\gamma$ was arbitrary this shows that $\alpha$ is an upper bound of $X$. However, since $\alpha <\beta$ this contradicts the definition of $\beta$ as being the least upper bound of $X$, according to which $\alpha \ge \beta$.

Since all cases lead to a contradiction it cannot be that $\beta =\sup <!--\; FUNCTION\; APPLICATION\; -->X$ is a successor and therefore by definition is a limit ordinal. □