# The comprehension axioms follows from the replacement schema.

I hope to show that the comprehension axioms follows from the replacement schema.

This is a solution that professor wrote.

$P(u,u)$: every set $u$, exists an unique $u$ such that $\psi(u)$.

Then exist a set $y=\{u:P(u,u) \text{ and } u \in X\}$

Let $F=\{(u,u) : \psi(u)\}$. Then $F(X) \in u$ and $y=\{u \in F(x) : \psi(u)\}$

I took some notes but I think that there are much errors.

• Maybe the answers here will be helpful: math.stackexchange.com/questions/32483/… – Apostolos Jun 15 '13 at 15:37
• Thanks but I don't know the separation axiom and I don't know the set theory well... It is too hard to me. – Guillermo Jun 15 '13 at 15:47
• The terms "Separation Schema" and "Comprehension Schema" are used pretty much interchangeably. Some texts differentiate between them (one being "bounded" and the other "unbounded"), but unless otherwise stated, it is usually safe to assume they are the same. – user642796 Jun 15 '13 at 15:57
• I have no idea why three people want to close this as not a real question; the question is obvious. – Brian M. Scott Jun 16 '13 at 6:37

Suppose that $x$ is a set, and $\varphi(u)$ is a formula. Let $\psi(u,v)$ be the formula $$u=v\land\varphi(u)\;.\tag{1}$$ The first conjunct, $u=v$, of $(1)$ ensures that for any $u$ there is at most one $v$ such that $\psi(u,v)$ holds, namely, $u$ itself. Thus, the formula $\psi$ behaves like a function, and the axiom schema of replacement says that if $z$ is any set, we can form the set

$$\left\{v:\exists u\in z\big(\psi(u,v)\big)\right\}\;.\tag{2}$$

If we write $v=F(u)$ as an abbreviation for the statement $\psi(u,v)$, we can rewrite $(2)$ in a more intuitive form: it’s the set

$$\{F(u):u\in z\}\;.$$

Now take for $z$ the specific set $x$, and expand $\psi(u,v)$. Then $(2)$ becomes

$$\left\{v:\exists u\in x\big(u=v\land\varphi(u)\big)\right\}\;,$$

which is easily seen to be the same as

$$\{u\in x:\varphi(x)\}\;.$$

Thus, we’ve deduced this specific instance of the axiom schema of comprehension from the axiom schema of replacement.

My notation is a little different, but this is essentially the argument that your professor gave.

• The version of the replacement axiom I've been using is this: $\forall x \in A \exists ! y \varphi(x,y) \rightarrow \exists B \forall x \in A \exists y \in B \varphi (x,y)$. With that I'm having difficulty seeing how the $\psi$ that you're using satisfies the axiom. Say for instance $\varphi(u)$ is false for some members of $A$. These would be counterexamples to $\forall x \in A \exists ! \varphi (x,y)$. – objectivesea Feb 18 '14 at 2:09
• I think, at the end of the proof, you mean $\{u\in x:\varphi(u)\}$ rather than $\{u\in x:\varphi(x)\}$. – Le Anh Dung Feb 20 '18 at 7:54