312: Well-Ordered Subset with Inclusion Ordering Is Chain in Base Set

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

A description/proof of that well-ordered subset with inclusion ordering is chain in base set

---

Topics

About: set

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of well-ordered set.
• The reader knows a definition of chain in set.

---

Target Context

• The reader will have a description and a proof of the proposition that for any set, any well-ordered subset with the inclusion ordering is a chain in the base set.

---

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 set, $S$, any subset, $S_1\subseteq S$, that is well-ordered by the inclusion ordering is a chain in $S$.

---

2: Proof

For any elements, $p_1,p_2\in S_1$, as $\{p_1,p_2\}$ is a subset of $S_1$, there is the smallest element in it, by the definition of well-ordered set. So, $p_1\subseteq p_2$ or $p_2\subseteq p_1$, which means that $S_1$ is a chain in $S$, by the definition of chain in set.

---

References

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