Axiom of Swelledness is implied by Axiom of Replacement
Theorem
Let the Axiom of Replacement (in the context of class theory) be accepted.
Then the Axiom of Swelledness holds.
Proof
Recall the Axiom of Replacement:
For every mapping $f$ and set $x$ in the domain of $f$, the image $f \sqbrk x$ is a set.
Symbolically:
- $\forall Y: \map {\text{Fnc}} Y \implies \forall x: \exists y: \forall u: u \in y \iff \exists v: \tuple {v, u} \in Y \land v \in x$
where:
- $\map {\text{Fnc}} X := \forall x, y, z: \tuple {x, y} \in X \land \tuple {x, z} \in X \implies y = z$
and the notation $\tuple {\cdot, \cdot}$ is understood to represent Kuratowski's formalization of ordered pairs.
Recall the Axiom of Swelledness:
- $V$ is a swelled class.
That is:
Let $x$ be a set.
Let $A$ be a class such that $A \subseteq x$.
Suppose $A$ is the empty class.
Then by the Axiom of the Empty Set, $A$ is a set.
Suppose $A$ is a non-empty class.
Let $c \in A$ be arbitrary.
Let $f$ be:
- the class of all ordered pairs $\tuple {a, a}$ for $a \in A$
along with:
- all ordered pairs $\tuple {y, c}$ where $y \in x \setminus A$.
Thus:
and:
- $\forall y \in x: \map f y = \begin {cases} y & : y \in A \\ c & : y \notin A \end {cases}$
Thus:
- $\forall y \in x: \map f y \in A$
and in fact:
- $f \sqbrk x = A$
We have by hypothesis that $x$ is a set.
Then by the Axiom of Replacement:
- $A$ is therefore also a set.
Hence the result.
$\blacksquare$
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 3$ The axiom of substitution