320: Ordinal Number Is Limit Ordinal Number iff It Is Nonzero and Is Union of Its All Members

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that ordinal number is limit ordinal number iff it is nonzero and is union of its all members

---

Topics

About: set

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

---

Starting Context

• The reader knows a definition of limit ordinal number.
• The reader admits the proposition that the inclusion relation is equivalent with the membership relation for the ordinal numbers collection.

---

Target Context

• The reader will have a description and a proof of the proposition that any ordinal number is a limit ordinal number if and only if it is nonzero and is the union of its all the members.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Description

For any ordinal number, $o$, $o$ is a limit ordinal number if and only if $o\ne 0$ and $o={\cup }_{o^{\prime }\in o}{o}^{\prime }$.

---

2: Proof

Let us suppose that $o\ne 0$ and $o={\cup }_{o^{\prime }\in o}{o}^{\prime }$.

If $o$ was a successor ordinal number, $o=o^{″+}$, $o^{″}\in o$ and $o^{″}\subset o$ by the proposition that the inclusion relation is equivalent with the membership relation for the ordinal numbers collection. For any $o^{\prime }\in o$, $o^{\prime }=o^{″}$ or $o^{\prime }\in o^{″}$ as $o^{″}$ is the largest smaller than $o$. So, by the proposition that the inclusion relation is equivalent with the membership relation for the ordinal numbers collection, $o^{\prime }\subseteq o^{″}\subset o$. So, ${\cup }_{o^{\prime }\in o}{o}^{\prime }\subset o$, a contradiction. So, $o$ is a limit ordinal number.

Let us suppose that $o$ is a limit ordinal number.

For any $o^{\prime }\in o$, by the proposition that the inclusion relation is equivalent with the membership relation for the ordinal numbers collection, $o^{\prime }\subset o$, so, ${\cup }_{o^{\prime }\in o}{o}^{\prime }\subseteq o$. For any $o^{″}\in o$, $o^{″+}\in o$, because $o^{″}\in o^{″+}$, $o^{″+}\ne o$, and $\mathrm{\neg }{o}^{″}\in o\in o^{″+}$. So, $o^{″}\in {\cup }_{o^{\prime }\in o}{o}^{\prime }$, because $o^{″}\in o^{″+}$ while $o^{″+}$ is an $o^{\prime }$. So, $o\subseteq {\cup }_{o^{\prime }\in o}{o}^{\prime }$. So, $o={\cup }_{o^{\prime }\in o}{o}^{\prime }$.

---

3: Note

If $o=0$, $o={\cup }_{o^{\prime }\in o}{o}^{\prime }$, because there is no $o^{\prime }$, and the union is supposed to be the empty set in that case, while $0$ is the empty set. So, the nonzero-ness condition is required.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>