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So let us say that there is time variable $t$ that can only be natural number. And then, for each $t$ there are data for each variable $a$, $b$, $c$, and so on. And then we have variable $y$. We want to express $y$ in term of $a$,$b$,$c$ and other variables. The equation does not have to be linear and it can be non-linear. In such case, assume that function only has to fit data points and for $t$ that has not been reached, things can be extrapolated.

  1. So if we have data points for each variable, then what would be the number of functions that would fit perfectly to form a function that maps from domain variables to $y$?

  2. what would be the minimum and maximum bound on the number of possible functions as data are added to every variable and therefore $t$ increases - not just when $t \to \infty$ but how this will change depending on $t$?

Every variable is real number variable.

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Fixing a real function at a countable number of points will not make a difference to the number of such functions. Or was the question something else? –  Mark Bennet Mar 30 at 15:32
    
So basically I am asking if we have data points for each variable, then what would be the number of functions that would fit perfectly to form a function that maps from domain variables to $y$. –  Cato Mar 30 at 15:37
    
Don't forget to mark an answer as correct or add a comment explaining further confusion –  Joshua Biderman Apr 2 at 14:37

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Even if you fix countably many such points, there are uncountably many such functions. You can even have uncountably many such continuous functions.

The function you are constructing sends countably many points in the domain to countably many points in the range. This leaves uncountably many points in the domain that can be assigned in any combination, so there are uncountably many ways to assign those points.

For continuity, it's a simply matter of noting that the countably many fixed range points partition $\mathbb{R}$ into open intervals, and it's a simple matter to find infinitely many continuous functions on an open interval with specified endpoints.

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Oh. But then why is it the case? Is it covered in real analysis textbook? –  Cato Mar 30 at 15:55
    
The function you are constructing sends countably many points in the domain to countably many points in the range. This leaves uncountably many points in the domain that can be assigned in any combination, so there are uncountably many ways to assign those points. –  Joshua Biderman Mar 30 at 16:02

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