Axiom schema of replacement
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\newcommand{\limplies}[0]{\rightarrow} \newcommand{\liff}[0]{\leftrightarrow}
Contents
[hide]Definition
Let \varphi(a,b,p) be a formula where a and b are free variables but p is a parameter (or a parameter pack) and describes a function between classes.
The axiom schema of replacement posits that if F is some class function then for all sets X, F(X) - the image of X under F - denoted F(X) is also a set[1].
We state it formally as follows:
- \Big(\underbrace{\forall x\forall y\forall z\big[\big(\varphi(x,y,p)\wedge\varphi(x,z,p)\big)\limplies y\eq z\big]}_{\text{if }(a,b)\in f\iff\varphi(a,b,p)\text{ then }f\text{ acts like a function relation} }\Big) \limplies \Big(\underbrace{\forall X\exists Y\forall y\big[y\in Y\liff\exists x[x\in X\wedge\varphi(x,y,p)]\big]}_{y\in Y\iff\text{the 'image' of }x\text{ under the 'function' is }y }\Big)[Note 1]
- By rewriting for-all and exists within set theory we can make a small change to the \exists x part:
- \Big(\forall x\forall y\forall z\big[\big(\varphi(x,y,p)\wedge\varphi(x,z,p)\big)\limplies y\eq z\big]\Big) \limplies \Big(\forall X\exists Y\forall y\big[y\in Y\liff\exists x\in X[\varphi(x,y,p)]\big]\Big)
- By rewriting for-all and exists within set theory we can make a small change to the \exists x part:
Notes
- Jump up ↑ Here it is without the underbraces:
- \Big(\forall x\forall y\forall z\big[\big(\varphi(x,y,p)\wedge\varphi(x,z,p)\big)\limplies y\eq z\big]\Big) \limplies \Big(\forall X\exists Y\forall y\big[y\in Y\liff\exists x[x\in X\wedge\varphi(x,y,p)]\big]\Big)