Axiom of Replacement

 Once we have ordered pairs (which we got in the last post), we are ready to define all kinds of important mathematical objects. The most obvious one is probably the product $A\times B$ of two sets $A$ and $B$. The definition is pretty straightforward:

$$A\times B:=\{⟨a,b⟩:a\in A,b\in B\}.$$

However, we should ask ourselves if this is a set: as far as we know it, there is nothing preventing this one from being a proper class! So, as you might've guessed, we solve our problems by declaring new axioms. 😎 I'll now try to describe Axiom of Replacement, which should (together with our old axioms) imply that $A\times B$ is a set.

Remember that we had the notion of a predicate $P(x)$. What we meant here was a property that a set may or may not have. Very similarly, we can consider a property $R(x,y)$ that the ordered pair of sets $x$ and $y$ may or may not have. One way to think about it is that we can look at some predicate $R^{\ast }(z)$ such that a set $z$ has this property only if it is of the form $⟨x,y⟩$ and in that case, we write $R(x,y)$ instead of $R^{\ast }(⟨x,y⟩)$. The point is simply that $R$ is some property of $x$ and $y$ together (and in that order). 

Suppose now that we have some set $A$ and some property $R(x,y)$. We would like to replace elements of $A$ according to $R$, but this probably needs some clarification. Let's say that $a\in A$. We then have this class:

$$R_a:=\{y:R(a,y)\}.$$

If for example $R(x,y)$ is true for all sets $x$ and $y$, then $R_a=V$ and we cannot expect to replace $a$ inside $A$ by all sets and still get a set-sized object! What I'm trying to say is that if we want to do replacing, we need some restrictions. Here's one: for all sets $a\in A$, there exists unique set $b$ (in the whole universe) such that $R(a,b)$ holds. This restriction seems promising because then for all $a\in A$, class $R_a$ is in fact a set with a single element, call it $r_a$. What we now mean by replacement is that we would like the class

$$\{r_a:a\in A\}$$

to also be a set: we replace every element of $A$ by some other element. Axiom of Replacement simply asserts that this is indeed true.

Okay, let's see why this axiom does the job: let's fix two sets $A$ and $B$ and let's show that $A\times B$ is also a set. For any element $a\in A$, we may replace each $b\in B$ by $⟨a,b⟩$. This leads to the set

$$C_a:=\{⟨a,b⟩:b\in B\}.$$

Now, each element $a\in A$ can be replaced by the set $C_a$, which spits out the set

$$D:=\{C_a:a\in A\}.$$

What are the elements of elements of $D$? An element of $D$ is of the form $C_a$ for some $a\in A$, while its elements are of the form $⟨a,b⟩$ for some $b\in B$. Well then, this just means that the set of the elements of elements of set $D$ is equal to $A\times B$, or in other words,

$$A\times B=\bigcup D.$$

If you recall Axiom of Union, you'll realize that, since $D$ is a set, then also $\bigcup D$ is a set, and we are done.