6 ms·

Do you mean: for each finite symbol set S, the set of all finite definitions using S is enumerable? That's clearly true. But then you're not really saying that

by gruseom 14y ago

Do you mean: for each finite symbol set S, the set of all finite definitions using S is enumerable? That's clearly true. But then you're not really saying that there are numbers with no possible definition, just that there is no particular finite symbol set S that can define them all. Or do you mean: the set of all finite definitions in any finite symbol set is enumerable? That's not clear. Are there countably many such symbol sets?

btilly 14y ago
What I mean is this. Given a finite symbol set, the set of all finite definitions is enumerable. And the enumeration is simple. We enumerate all possible definitions of length 1, 2, 3, 4, etc. We order them first by length, and then by alphabetical order. We strike out all definitions that are not definitions, or define a number already defined. This leaves us with an enumeration of all numbers definable with that set of symbols. (Note, a constructivist will insist strenuously that this is not actually an enumeration.) Given that there are uncountably many real numbers, uncountably many of them are not in that enumerable set, and therefore almost all real numbers cannot be defined using that finite set of symbols. The possibility of uncountably many possible symbol sets with a similar plethora of possible associated definitions is irrelevant, because any potential symbol set that cannot ultimately be described in the finite alphabet that we use with the language English is irrelevant to us.
- phaedrus 14y ago
  As a programmer I find constructivism is my default position if I think of math intuitively; perhaps that's why I butted heads so much with many of my math teachers. But, your argument actually makes me less sure of my agreement with it: it seems clear that at bottom constructivism must lead to the rejection of the concept of continuously variable quantities. It is the horns of a dilemma: if you reject Cantor you reject the existence of "analog" as even an admissible concept, on the other hand if you accept continuous variability then you must also accept the shocking conclusion that the overwhelming majority of the set you assert you believe exists (the reals) cannot even be described or calculated.
  - btilly 14y ago
    But constructivism does not give up continuity. All of the traditional definitions carry over. What you do have to give up, though, are discontinuous functions on continuous sets. They are not well defined for numbers that are the result of an infinite sequence converging near the border of the discontinuity. You can still give them a formal treatment, of course, but they aren't really functions. However if you're attached to your step functions, never fear. It would not be the first function-like thing that isn't a function. For another example consider the well-known derivative of the step function, the Dirac delta. Not actually a function in classical mathematics, though it is certainly useful.

by gruseom 14y ago

btilly 14y ago

What I mean is this. Given a finite symbol set, the set of all finite definitions is enumerable. And the enumeration is simple. We enumerate all possible definitions of length 1, 2, 3, 4, etc. We order them first by length, and then by alphabetical order. We strike out all definitions that are not definitions, or define a number already defined. This leaves us with an enumeration of all numbers definable with that set of symbols. (Note, a constructivist will insist strenuously that this is not actually an enumeration.) Given that there are uncountably many real numbers, uncountably many of them are not in that enumerable set, and therefore almost all real numbers cannot be defined using that finite set of symbols. The possibility of uncountably many possible symbol sets with a similar plethora of possible associated definitions is irrelevant, because any potential symbol set that cannot ultimately be described in the finite alphabet that we use with the language English is irrelevant to us.

phaedrus 14y ago
As a programmer I find constructivism is my default position if I think of math intuitively; perhaps that's why I butted heads so much with many of my math teachers. But, your argument actually makes me less sure of my agreement with it: it seems clear that at bottom constructivism must lead to the rejection of the concept of continuously variable quantities. It is the horns of a dilemma: if you reject Cantor you reject the existence of "analog" as even an admissible concept, on the other hand if you accept continuous variability then you must also accept the shocking conclusion that the overwhelming majority of the set you assert you believe exists (the reals) cannot even be described or calculated.
- btilly 14y ago
  But constructivism does not give up continuity. All of the traditional definitions carry over. What you do have to give up, though, are discontinuous functions on continuous sets. They are not well defined for numbers that are the result of an infinite sequence converging near the border of the discontinuity. You can still give them a formal treatment, of course, but they aren't really functions. However if you're attached to your step functions, never fear. It would not be the first function-like thing that isn't a function. For another example consider the well-known derivative of the step function, the Dirac delta. Not actually a function in classical mathematics, though it is certainly useful.