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In axiomatic set theory a function $F$ is defined to be a class/set all of whose members are ordered pairs. Given a function $F$ there are set theoretic equations involving union, intersection and difference for calculating the domain, range, and also given a set $x$ the value $F(x)$ of $F$ at $x$.

I can work out the equation for the domain and range based on the Kuratowski definition of an ordered pair $<x,y>=\{\{x\},\{x,y\}\}$ namely

$$Dom(F)=\bigcup\{\bigcap P\mid P\in F\}$$ $$Ran(F)=\bigcup\{\bigcup P\setminus\bigcap P\mid P\in F\}$$

But I can't find the formula for $F(x)$ anywhere! Can someone please provide it.

Thanks

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Note that axiomatic set theory is usually formalized in a logical language where there are no set-valued expressions other than variables. Whenever something that looks like a set-valued expression seems to appear in a formula, it is actually to be understood as an abbreviation of a more complex logical formula that intuitively claim that some variable $x$ happens to equal what we want to formula to evaluate to.

This means that whenever we introduce an expression-like notation, for example $\{x,y\}$, what we actually need to define is a formula that represents the proposition $z=\{x,y\}$, namely in this case $$ z=\{x,y\} \quad\text{means}\quad \forall w(w\in z\leftrightarrow (w=x\lor w=y)) $$ We can then prove (using the axiom of pairing) that $$ \forall x\forall y\exists! z\, (z=\{x,y\}) $$ and once that is known, a standard process allows us to unfold formulas involving $\{\cdot,\cdot\}$ into formulas built only from the predicates $=$ and $\in$.

Once we know that this is how you define notation, seeing how to define function application becomes easy. Namely, what we need to define is not the term $F(x)$, but the assertion $y=F(x)$. This has a quick solution: $$ y=F(x) \quad\text{means}\quad \left<x,y\right> \in F $$ which however only works if $F$ is known to be a function and $x$ to be in its domain. If we want $F(x)$ to be meaningful always (even though we don't care about what precisely it means when $x$ is not in the domain of $F$), we could write, for example $$ y=F(x) \quad\text{means}\quad y = \cup\{z\in \cup{\cup} F \mid \left<x,z\right> \in F \} $$ which works correctly when $x\in\operatorname{Dom} F$ and produces some definite set otherwise.

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  • $\begingroup$ Thanks, it was that last formula I was looking for. Appreciate your views on axiomatic set theory as well, but as a matter of taste I prefer to keep the intrusion of first order language to a minimum (axiom of comprehension plus essential definitions) then use a more algebraic approach. I think it is easier with Morse-Kelley than ZF set theories because with ZF or even NGB set theory the language intrudes because of how those theories are forced to treat classes. $\endgroup$ – Mark Kortink Jul 25 '16 at 4:38

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