📂Set Theory

Substitution Axiom Form

Axiom

$$\forall X \left( \forall x \in X \exists ! y \left( p(x,y) \right) \implies \exists Y \forall x \in X \exists y \in Y \left( p(x,y) \right) \right)$$ The range exists for all functions.

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• The symbol $\exists !$ signifies uniqueness.
• Here, $p(x,y)$ is a propositional function in $X \times Y$.

Explanation

Although propositional function $p(x,y)$ is a function, strictly speaking, it has not yet been defined as a function, and even if it were defined as a function, it is not the function itself mentioned in the axiom above. What logical expression $p(x,y)$ conveys is that when $x \in X$ is given, $y \in Y$ exists:

• For instance, $p$ can be given as $p(x,y): y = 2x$, and in this case, the function that has a range is $f(x) = 2x$, not $p(x,y)$. It is the image $Y = f [ 0 ,1] = [0,2]$ of $X = [0,1]$ for the function $f$ that exists according to the substitution axiom form.

The reason why it is called an axiom form rather than an axiom is because this axiom exists innumerably for innumerably many $p(x,y)$. If there are two different propositional functions, $p_{1}(x,y): y = \sin x$ and $p_{2}(x,y): y = 2x$, the existence of $Y = f_{2} [ 0,1] = [0,2]$ is guaranteed not by the ‘substitution axiom for $p_{1}(x,y)$’ but by the ‘substitution axiom for $p_{2}(x,y)$’.