I'm trying to construct a rigorous set theoretic formulation of the old formulation of functions in terms of variables. I began with the following old school definition of variable:

Definition XI: A variable over a set S is an unspecified individual element of a set. The set S is called the range of the variable, any specified element of the set S is called a value of the variable.

This leads to the following definition of function given by Arnold Dresden:

Definition XVI: When the two variables x and y are so related that to each value of the range of x there corresponds a value of y, we say that y is a function of x. Notation: y= f (x).

This is more or less the old Dirichet definition of function given in the slightly more precise language of variables.

Here's my proposed "rigorous" definition of variables:

Def: Let x be a symbol. Let S be a nonempty set where x can be assigned any member of S as it's value. Then we define the variable U with range S as follows: U = {x} X S = { (x,a) : $a \in S$} Note U is a relation in the set theoretic sense since every ordered pair has the same first member, namely x.

I'm assuming the Kuratowski definition of an ordered pair i.e. (x,a) = { {x}, {x,a}}

Now here's where I need someone to double check me. Let U be the variable whose range is the domain S of a function $f:S\rightarrow T$ and let V be the variable whose range is the functions's range. U = {x} X S = { (x,a) : $a \in S$} V = {y} X T = { (y,b) : $a \in T$}

I want to now define a function between variables x on the domain and the range y using these definitions of variables.

Here's my question: How do I calculate U X V?

Here's my computation, which I'm pretty sure is wrong:

U X V = ( {x} X S) X ({y} X T) = (x,y) X (S X T).

Something looks VERY wrong here. Please someone take another look.


1 Answer 1


{x}×S × {y}×T = { ((x,s), (y,t)) : s in S, t in T }

  • $\begingroup$ That was my initial computation, but I was a bit confused. We simply end up with a set of ordered pairs that's the union of the variables U and V? If so, then the definition of f is very simple: It's the subset of U X V where no 2 different ordered pairs have the same first member i.e. (($x_1$, $s_1$), (y,t)) = (($x_2$, $s_2$), (y,t)) iff $x_1$ =$x_2$ AND $s_1$ = $s_2$. Which is relatively simple if one wants to continue using the vocabulary of variables.in modern set theoretic language, but not as simple as working directly with set constructions. $\endgroup$ Nov 18, 2019 at 0:07
  • $\begingroup$ if this is correct,then the problem was I'm overthinking this. We just build each ordered pair in U X V by taking an ordered pair of U as the first member and the "corresponding" ordered pair of V as the second member. $\endgroup$ Nov 18, 2019 at 0:13

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