3 ms·

> That is the definition of a function, but it's purely a formalism... Isn’t the function definition meant to be interpreted as “maps consistently to exactly o

> That is the definition of a function, but it's purely a formalism... Isn’t the function definition meant to be interpreted as “maps consistently to exactly one element of the codomain”? So an ordered pair of R^2 is still one element of R^2 It seems the parent has just mixed up the domain and codomain, because under that assumption he would be right about both the definition and bijectivity

thaumasiotes 3y ago
I can't tell what you mean by your emphasis on the word "consistently". Functions are not stochastic; f(y) is f(y) regardless of how many times you ask what f(y) is. The formal definition guarantees that whenever a = b, f(a) = f(b). You use it when you need that guarantee. An ordered pair drawn from ℝ² is in some sense a single value. In another sense, it is two values. Which way you want to think about it depends on what you're going to do with it; if you're thinking about square roots of real numbers, it will be more useful to think of it as two values. > It seems the parent has just mixed up the domain and codomain, because under that assumption he would be right about both the definition and bijectivity He still wouldn't be right about bijectivity; you also need the assumption that a function is defined over its entire domain.
- lying4fun 3y ago
  >is in some sense a single value. In another sense, it is two values. What I tryed to say is that the sense in which “exactly one element” is used in the definition of function is inclusive of codomain being R^n, so it confused me why you would provide a function that has a codomain of R^2 as something that suggests deviation from the formalism. It just seemed misleading to phrase it that way I thought consistently would convey the idea I had, nevermind if it doesn’t My bad about bijectivity, I see it now, you’re right

thaumasiotes 3y ago

I can't tell what you mean by your emphasis on the word "consistently". Functions are not stochastic; f(y) is f(y) regardless of how many times you ask what f(y) is. The formal definition guarantees that whenever a = b, f(a) = f(b). You use it when you need that guarantee. An ordered pair drawn from ℝ² is in some sense a single value. In another sense, it is two values. Which way you want to think about it depends on what you're going to do with it; if you're thinking about square roots of real numbers, it will be more useful to think of it as two values. > It seems the parent has just mixed up the domain and codomain, because under that assumption he would be right about both the definition and bijectivity He still wouldn't be right about bijectivity; you also need the assumption that a function is defined over its entire domain.

lying4fun 3y ago
>is in some sense a single value. In another sense, it is two values. What I tryed to say is that the sense in which “exactly one element” is used in the definition of function is inclusive of codomain being R^n, so it confused me why you would provide a function that has a codomain of R^2 as something that suggests deviation from the formalism. It just seemed misleading to phrase it that way I thought consistently would convey the idea I had, nevermind if it doesn’t My bad about bijectivity, I see it now, you’re right