To me its not unlikely to have such equation which you may put any value for x from 0 to 1, and get a point on the line y=0 if the point wasn't on Cantor set, or on y=1 if the point was.
As such there is no algebraic equation. However, a procedure based on construction of the Cantor set can be designed to check if the number is on the set or not. For example, consider x in [0,1]. Represent it in ternary fraction. If it contains digits 0 and 2 but no 1 at all, then it is on the Cantor set. If somewhere digit 1 comes it is not on the Cantor set.
There is no algebraic equation to cover all cantor set points in [0,1]. Only the union of the functions of construction of cantor set may cover the all cantor set points in [0,1] othwise it is not.
One can construct a function f : [0,1] ---> R following the description given above by R. C. Mittal. For x in [0,1] consider the binary expression which does not end with an infinity of ones (there always exist such an expression!). Then replace the ones with twos and reconsider the infinite expression as a ternary expression. You have got a point of the Cantor set. This function has as image the Cantor set, is monotone increasing, but is not continuous - because as such it should carry intervals in intervals. Of course, no such continuous function does exist.
Yes, a polynomial equation is what I'm looking for, but then if there is no such thing, how can we prove it?
Start with the fact that polynomial of degree n has exactly n roots (counted by multiplicity).
Or use the fact that polynomials are continuous, and the graph of the Cantor set is not.
There is a procedure to construct Cantor sets using trenary system but no an alebraic equation.
Consider a polynomial expression f(x,y) = 0 such that if x in R \ C, f(x,0) = 0 and if x in C, f(x,1) = 0. C being infinite, there must exist a point (x, 1) with x in C such that the partial derivatives of f in this point are not 0. By the local inversion theorem, the set {(x,y) | f(x,y) = 0} must contain an arc of curve going through (x,1). But this is a contradiction, because C is totally disconnected. So no such polynomial does exist.
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