Axiom of Swelledness is implied by Axiom of Replacement

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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:

Every subclass of a set is a set.


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:

$f$ is a mapping whose domain is $x$

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

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