Nesthood has Finite Character
Theorem
Let $P$ be the property of sets defined as:
- $\forall x: \map P x$ denotes that $x$ is a nest.
Then $P$ is of finite character.
That is:
- $x$ is a nest
- every finite subset of $x$ is a nest.
Proof 1
By definition, a nest $N$ is a class on which $\subseteq$ is a total ordering.
Here we are given that $N$ is a set.
The result follows from Property of being Totally Ordered is of Finite Character.
$\blacksquare$
Proof 2
By definition, a nest is a class on which $\subseteq$ is a total ordering.
Sufficient Condition
Let $x$ be a nest.
Let $y \subseteq x$.
From Restriction of Total Ordering is Total Ordering it follows that $y$ is also a nest.
This holds in particular if $y$ is a finite set.
Hence the result.
$\Box$
Necessary Condition
Suppose that every finite subset of $x$ is a nest.
We have that Subset Relation is Ordering.
It remains to show that $\subseteq$ is total.
Let $y, y' \in x$ be arbitrary.
Since $\set {y, y'}$ is a finite subset of $x$, it is a nest.
Therefore, either $y \subseteq y'$ or $y' \subseteq y$.
Since $y, y'$ were arbitrary, it follows that $x$ is also a nest.
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles