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This might seem like a weird question, but is it actually possible to define a function for all possible inputs? By this, I really mean /all/ possible inputs, including numbers, true and false, sets, sets of sets, other functions, itself---everything. To me, this doesn't seem problematic, but maybe there's some kind of subtle reason why this cannot be done. Here's an example (maybe) of a function defined for all possible inputs:

$F(x) = \mathbf{true} \text{, if } x = 0 \\ F(x) = \mathbf{true} \text{, if } x = 1 \\ F(x) = \mathbf{false} \text{, otherwise.}$

Is there anything wrong with that?

enigmaticPhysicist
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  • I am starting to believe that your function might not be a mathematical function and your input not an element of some domain set, but that you mean a machine function and input is the input according to your machine model. Which would leave the question which machine model. – mvw May 10 '15 at 03:13
  • The identity function is defined for all possible inputs. – MJD May 10 '15 at 05:46
  • You can compare with Frege's view about predication : "a simple predication is analyzed in terms of falling under a concept, which in turn, is analyzed in terms of functions which map their arguments to truth values. By contrast, in the modern predicate calculus, this last step of analyzing predication in terms of functions is not assumed; predication is seen as more fundamental than functional application." For Frege, a concept is a (unary) function with an object as argument and a truth-value as value. 1/2 – Mauro ALLEGRANZA May 10 '15 at 12:19
  • Interestingly, his conception gives rise to Russell's Paradox. 2/2 – Mauro ALLEGRANZA May 10 '15 at 12:20
  • You might have know it, but this won't work in any theory such as ZFC that does not support the concept of a set of all sets: that is, you seem to want pick up a $F$ such that $F \in {0,1}^U$ and you want it to be a function of domain U and Boolean co-domain, where U denotes the set of all sets. Hence you need that $F \subseteq U \times {0,1}$, which is undefined, since U is not a set. – Bruno Bentzen May 10 '15 at 15:34
  • Oh, yeah, definitely won't work with ZFC. But see my comment in the first answer about a new definition of set (maybe) that would allow for a set of sets,and that afaict has no drawbacks. – enigmaticPhysicist May 10 '15 at 15:47

3 Answers3

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A function is normally understood to have domain that is a set of some kind. Doing so allows us to handle functions as sets themselves, namely sets of ordered pairs. For example, $f(x)=x^2$ on the integers can be thought of as $\{(1,1), (2,4), (-2,4),\ldots\}$.

However not all mathematical objects are sets; for a famous example consider Cantor's paradox, i.e. if you take all sets, the result is not a set. This object can be called a class, and such objects are so large and weird that we can't really define functions on them in the usual way. We can define function-like things, as done in the OP. However if we allow the domain to not be a set, then the resulting function is not a set. Thus it lives outside of familiar set theory. This makes it very unusual, since (apart from logicians) most people live their entire mathematical lives using objects that are defined within a set theory, typically ZFC or something similar.

vadim123
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  • Funny you should mention requiring the domain to be a set. I was actually going to use the idea of a function defined on all inputs to try to form a new definition of a set, as a way around Russell's paradox. Define a set S as a function which must be defined for all possible inputs. x is in the set S iff S(x) = true. Russell's paradox tries to construct a set R of all sets not containing themselves. As a function, R(x) = false if x is not a set, and if x is a set, R(x) = !(x(x)). This leads to R not being defined at x = R. Meaning R is not a set anyway, so there is no paradox. – enigmaticPhysicist May 10 '15 at 03:39
  • Such a definition for a set would also trivially resolve Cantor's paradox, since it is very easy to construct a function whose output is true for all inputs. Interpreted as a set, this set would contain everything. With this definition of set, a function could still be interpreted as a set of ordered pairs, even if the function is defined for all possible inputs. – enigmaticPhysicist May 10 '15 at 03:42
  • @enigmaticPhysicist Do you think that Russell's paradox and Cantor's paradox have not already been resolved? An interesting topic to investigate might be to compare your definition of "set" with the ZFC definition of "set." – David K May 10 '15 at 12:04
  • I know ZFC resolves them, but so would the definition of set I just mentioned, but mine allows for a set of all sets. – enigmaticPhysicist May 10 '15 at 15:49
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Your $F$ is actually a logical predicate:

$\forall x:[F(x)\iff x=0\lor x=1]$

In mathematics (if maybe not in philosophy), you would probably want to restrict the domain of quantification (e.g. to the set of natural numbers $\mathbb{N}$) as follows:

$\forall x\in \mathbb{N}:[F(x)\iff x=0\lor x=1]$

Dan Christensen
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Your approach is fine as long as you know how $0$ and $1$ are represented. An even simpler approach is $F(x)=2$ where you ignore the input entirely. All that you need for a function is a unique response to every input in the domain. Why pay attention to the input?

Ross Millikan
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