how to prove this result.

$\alpha$ is limit ordinal if and only if $\beta<\alpha$ implies $\beta+1<\alpha$ for any $\beta$.

I am getting confused because of the definition of limit ordinal. it says

$\alpha=\sup\{\beta: \beta<\alpha\}=\bigcup\alpha$

for sup I am ok but for $\alpha=\bigcup\alpha$ how ??


To show $\alpha = \bigcup \alpha$ implies ($\beta < \alpha$ implies $\beta + 1 < \alpha$) assume that $\alpha = \bigcup \alpha$ and not ($\beta < \alpha$ implies $\beta + 1 < \alpha$):

If $\beta < \alpha$ and $\beta + 1 \ge \alpha$ then $\beta < \alpha \le \beta + 1$. Since there are no ordinals between $\beta$ and $\beta + 1$ it follows that $\alpha = \beta + 1$. But then $\alpha = \bigcup \alpha = \bigcup \beta + 1 = \beta < \alpha$ which is a contradiction.

For the other direction, assume ($\beta < \alpha$ implies $\beta + 1 < \alpha$). The inequality $\bigcup \alpha \le \alpha$ always holds. To show $\alpha \subseteq \bigcup \alpha$ let $\beta \in \alpha$. Then by assumption $\beta + 1 \in \alpha$. Since $\beta \in \beta + 1$ it follows that $\beta \in \bigcup \alpha$.

  • $\begingroup$ Note how you can exchange $\le$ and $\subseteq$ and $\in$ and $<$ because the sets involved are ordinals. I hope I understood your question correctly. $\endgroup$ Apr 6 '13 at 15:21
  • $\begingroup$ It is ok that's what I want, but the last statement is not clear " $\beta\in\bigcup\alpha$ $\endgroup$ Apr 6 '13 at 15:27
  • $\begingroup$ @ZirkoTammado The last statement is a consequence of $\beta + 1 \in \alpha$. Then $\bigcup \alpha = \beta + 1 \cup \dots$ and since $\beta \in \beta + 1$ then also $\beta \in \bigcup \alpha$. $\endgroup$ Apr 6 '13 at 15:30
  • $\begingroup$ I know what does this mean but we want to show that $\beta+1\in\bigcup\alpha=\alpha$ $\endgroup$ Apr 6 '13 at 15:35
  • $\begingroup$ @ZirkoTammado Apply the assumption to $\beta + 2$ to get $\beta + 2 \in \alpha$. $\endgroup$ Apr 6 '13 at 15:37

Recall that ordinals are sets. $\bigcup\alpha$ is the union of all the members of $\alpha$. That is, $x\in\bigcup\alpha$ if and only if there is some $\beta\in\alpha$ such that $x\in\beta$.

Recall that ordinals are transitive sets. It turns out that if $A$ is a set of ordinals then $\sup A=\bigcup A$.

Recall that ordinals are sets of ordinals, and every ordinal is the set of those ordinals strictly smaller than itself. Therefore $\sup\{\beta:\beta<\alpha\}=\bigcup\alpha$.

Here is an outline of the equivalence,

If $\alpha$ is not a limit ordinal then $\alpha=\beta+1$ for some $\beta$, therefore $\sup\{\gamma:\gamma<\alpha\}=\beta$, because $\beta$ is in fact the maximum of that set. Therefore if $\alpha$ is not a limit ordinal, $\alpha\neq\sup\{\beta:\beta<\alpha\}$.

If there is $\beta<\alpha$ such that $\beta+1\nless\alpha$, then by the fact ordinals are linearly ordered $\alpha\leq\beta+1$, and therefore $\alpha=\beta+1$. It follows that $\{\beta:\beta<\alpha\}$ has a maximal element, $\beta$.

(Some minor details and small arguments might need to be added, but the idea is there.)

  • $\begingroup$ Excellent Asaf for this nice explanation. Could you please give me a proof of the above result as well. $\endgroup$ Apr 6 '13 at 15:17
  • $\begingroup$ @Zirko: Which one? $\sup A=\bigcup A$? $\endgroup$
    – Asaf Karagila
    Apr 6 '13 at 15:18
  • $\begingroup$ $\alpha$ is limit ordinal if and only if $\beta<\alpha$ implies $\beta+1<\alpha$ for any $\beta$. $\endgroup$ Apr 6 '13 at 15:20
  • $\begingroup$ Thanks alot I will try to catch it. You are always the best. $\endgroup$ Apr 6 '13 at 15:29

For each ordinal $\alpha$, the set $\beta:=\bigcup \alpha$ is also an ordinal. As it does not contain $\alpha$ itself it is either $=\alpha$ or $<\alpha$. The case $\beta<\alpha$ occurs iff $\alpha=\beta+1$, why?


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