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I have read two variations of Axiom Schema of Replacement:

  1. The first one is from https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory#6._Axiom_schema_of_replacement, and is stated as:

    $\forall \vec{w} \forall A [ \forall x (x\in A \implies\exists ! y \varphi(x,y,\vec{w}, A)) \implies \exists B \forall x (x \in A \implies \exists y (y \in B \land \varphi(x,y,\vec{w}, A)))]$

  2. The second one is from Definability and the Separation and Replacement Axiom Schemata, and is stated as:

    $\forall \vec{w} \forall A ( (\forall x \in A \exists ! y \varphi(x,y,\vec{w}, A) \implies (\exists B \forall y (y \in B \iff \exists x \in A \varphi(x,y,\vec{w}, A))))$

I would like to ask a few questions:

  1. Why we use relation $\color{red}{\implies}$ in formula $\exists B \forall x (x \in A \color{red}{\implies} \exists y (y \in B \land \varphi(x,y,\vec{w}, A)))$, but relation $\color{blue}{\iff}$ in formula $(\exists B \forall y (y \in B \color{blue}{\iff} \exists x \in A \varphi(x,y,\vec{w}, A))))$.

  2. I know $\color{red}{\vec{w}}$ is the vector of parameters in $\varphi(x,y,\color{red}{\vec{w}}, A)$. But why do we include $\color{blue}{A}$ in $\varphi(x,y,\vec{w}, \color{blue}{A})$ while $x$ is already in $\varphi(x,y,\vec{w}, \color{blue}{A})$? So what is the role of $\color{blue}{A}$ in $\varphi(x,y,\vec{w}, \color{blue}{A})$?

  3. Is it TRUE that set $B$ from the $\color{red}{first}$ one is the $\color{red}{CODOMAIN}$ of the function defined by $\varphi(x,y,\vec{w}, A)$, while set $B$ from the $\color{blue}{second}$ one is exactly the $\color{blue}{RANGE}$ or $\color{blue}{IMAGE}$ of the function defined by $\varphi(x,y,\vec{w}, A)$?

  4. Is it WRONG to use symbols $\color{red}{\exists ! y}$ ? Since $\color{red}{\exists ! y}$ means $\color{red}{\text{there exists only one y}}$, while Axiom schema of replacement states that $\color{blue}{\text{there's AT MOST one y}}$?

  5. If there is any other differences between these two variations, please let me know!

Many thanks for your help!

Community
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Akira
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1 Answers1

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The question comes down to what are your overall axioms.

If you include Separation, then the weaker form of Replacement, requiring only that there is some set which includes all the image is enough. Because we can separate the image from this set, obviously.

If you do not include Separation, then the weaker form is not enough if you want to conclude the image is a set.

Similarly with the question on the quantifiers over $y$. In principle, $\exists!y$ is the right thing, since we want to state that $\varphi$ defines a function when considering $A$ as a domain. But even if you require only "at most one", if you have Separation, then you can separate the actual domain from $A$ and work with that.

Asaf Karagila
  • 393,674
  • You have addressed Question 3 and 4. Could you please answer question 2 above? – Akira Dec 23 '17 at 10:48
  • Of these two variations, which one you are referring as the weaker form of Replacement? From your answer, I infer that we CAN NOT use the WEAKER form of Replacement to deduce axiom of separation, is my understanding correct? – Akira Dec 23 '17 at 10:51
  • Of these two formulations one clearly implies the other. And why are you USING weird CAPITALIZATION? It's weird. YOU know? – Asaf Karagila Dec 23 '17 at 12:53
  • I use CAPITALIZATION and color just to emphasize ^^. I can not understand why $\exists B \forall x (x \in A \implies \exists y (y \in B \land \varphi(x,y,\vec{w}, A)))$ implies $\exists B \forall y (y \in B \iff \exists x \in A \varphi(x,y,\vec{w}, A))$ and vice versa. Please elaborate on how to prove these two formulas are equivalent ! – Akira Dec 23 '17 at 13:34
  • What's wrong with boldface, italics or *both*? WHY do you have TO do THIS??? (Not to mention that as a color blind, the use of color is more confusing and distracting than helpful). The answer is that you can use Separation to separate the $B$ in the second axiom from the $B$ in the first axiom. – Asaf Karagila Dec 23 '17 at 13:52
  • I'm sorry for any inconvenience and confusion! From my understanding of your comment, the first axiom together with axiom of separation implies the second axiom. Furthermore, we can not deduce axiom of separation from the first axiom. Please check if my understanding is correct! – Akira Dec 23 '17 at 14:10
  • Yes, that is what I meant. – Asaf Karagila Dec 23 '17 at 16:08
  • Many thanks for your patient! – Akira Dec 23 '17 at 16:52
  • My last question in this topic is that: is it true that the first axiom is weaker than the second one? – Akira Dec 24 '17 at 01:58
  • Well, again, what would "weaker" mean here? With other axioms, they are equivalent. But without Separation, well, I don't quite remember, but I think there is some results on this. You can also read about the Collection schema, which is equivalent to Replacement when considering ZF, but is stronger when working in weaker systems. – Asaf Karagila Dec 25 '17 at 12:21