Tao Analysis I - 3.1.11

Exercise 3.1.11

Show that the axiom of replacement implies the axiom of specification.

Let's write out these axioms.

Axiom 3.7 (Replacement). Let $A$ be a set. For any object $x \in A$, and any object $y$, suppose we have a statement $P(x, y)$ pertaining to $x$ and $y$, such that for each $x \in A$ there is at most one $y$ for which $P(x, y)$ is true. Then there exists a set $\{y : P(x, y) \text{ is true for some } x \in A\}$, such that for any object $z$,

$$z \in \{y : P(x, y) \text{ is true for some } x \in A\} \iff P(x,z) \text{ is true for some } x ∈ A$$

It is worth noting that this axiom about the existence of the set $\{y\}$.

Axiom 3.6 (Axiom of specification). Let $A$ be a set, and for each $x ∈ A$, let $P(x)$ be a property pertaining to $x$ (i.e., for each $x \in A$, $P(x)$ is either a true statement or a false statement). Then there exists a set, called $\{x \in A : P(x) \text{ is true }\}$ (or simply $\{x \in A : P(x)\}$ for short), whose elements are precisely the elements $x$ in $A$ for which $P(x)$ is true. In other words, for any object $y$,

$$y \in \{x \in A:P(x) \text{ is true }\} \iff (y \in A \text { and } P(y) \text{ is true })$$

Again, this is an axiom about the existence of the set $\{y\}$.

Let's consider the Axiom of Replacement, and define the statement $P(x,y)$ as follows:

$$P(x,y) := (x=y) \land (Q(y) \text{ is true })$$

This definition of $P(x,y)$ also satisfies the axiom's requirement that for each $x \in A$ there is at most one $y$ for which $P(x,y)$ is true. This is because $x=y$.

The axiom tells us there exists a set $\{y : P(x, y) \text{ is true for some } x \in A\}$. We can write $P$ in terms of $Q$,

$$\begin{align}\{y : P(x, y) \text{ is true for some } x \in A\} &= \{y : x=y \land Q(y) \text{ is true for some } x \in A\} \\ \\ &= \{y : Q(y) \text{ is true for some } y \in A\}  \\ \\ &= \{y \in A: Q(y) \text{ is true }\}  \end{align}$$

The RHS is equivalent to the Axiom of Specification. Thus we have shown the axiom of replacement implies the axiom of specification. $\square$