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From axiom of replacement: I have given a set $S$ and have relation $R(x,y)$ such that $x \in S$, then there exist unique $y \in T$ which is image of set $S$ where $T$ is also a set.

But I can use Axiom schema of comprehension where I have $P(y)$ as predicate, then there exist a subset $T \subset S$ such that it consist of exactly those $y \in S $ for which $P(y)$ is true. In symbols I have: $$T = \{y\in S|P(y)\}$$ Then it says that axiom schema of comprehension is a consequence of axiom of replacement as:

$\bullet\,$ $\,\lnot(\exists y\in S : P(y))$ then define $T = \phi$

$\bullet\,$ $\, \exists \,\hat{y}\in S : P(y)$ then by definition I have $$\text{when } P(x) \text{ is true, then } R(x)=y$$ $$\text{when } P(x) \text{ is false, then } R(\hat{y})=y$$ From all of this $T$ is image is image of set $S$ under relation $R(x,y)$

My confusion is:

$ \color{red}{\text{ how }y\in S \text{ and axiom schema of comprehension is a consequence of axiom of replacement}}$ Please provide me one example also. Thanks in advance!

note: I searched on internet and it didn't help and I am physics person and started learning set theory.

rahul yadav
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1 Answers1

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Long comment

See Relation to the axiom schema of replacement for the derivation of Separation from Replacement.

Please, note that in the Axiom Schema of Replacement we use a formula $\varphi$ that is functional (which relates each set $x$ to a unique set $y$) and not a relation $R$ (i.e. a set).

For a relation $R$, i.e. a subset of $S \times T$, we have the domain $S$ and the codomain $T$. We define the range (or: image) of $R$ as the subset of the codomain:

$\text{Ran}(R) = \{ y \in T \mid R(x,y) \text { for some } x \in S \}$.

Thus, by definition: $\text{Ran}(R) \subseteq T$.

Mauro ALLEGRANZA
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  • $R(x, y)$ be $x = y\wedge P(x)$. $T ={(x,x) :\phi(x)} ={(x,y) :R(x,y)}$. $R(x,y)$ is a functional formula. Therefore, we have that The Separation Axioms follow from the Replacement Schema. $T$ $=$ ${y : (\exists x\in X)R(x, y)}$ $=$ ${y:(\exists x\in S)x = y\wedge P(x)}$ $=$ ${x:(\exists x\in S)\phi(x)}$ $=$ ${x\in S :P (x)}$. – rahul yadav Aug 14 '21 at 23:15
  • i am not getting how T is image – rahul yadav Aug 14 '21 at 23:20
  • This is a functional formula (i.e. for every u there is at most a single v for which φ(u,v,p) holds) and it is easy to verify that the image of φ(u,v,a) is indeed $${x\in a\mid \phi(x)}$$ This is from : https://math.stackexchange.com/a/32493/956732 – rahul yadav Aug 15 '21 at 00:11