[shoo]

> He works with some family of polynomial functions who agree on the sets K[a] that have open interior (2.1). Of course, two polynomials that agree on infinitely many points are identical.

Consider f(x, y) := xy and g(x, y) := xy^2

Fixing x=0 note that f and g agree along {(x, y) | x = 0, y in R}. But f is not identical to g. There is no open subset of R^2 such that f and g agree throughout the subset.

Would rewording "two polynomials that agree on infinitely many points are identical" as "two polynomials that agree on any open set" fix this? Or restrict the statement to polynomials in one variable only?

[m00n]

You're right, I was thinking about univariate polynomials. The "preprint" is only concerned with functions of one complex argument, so this should suffice.

[wbl]

C is not R^2. Neither f nor g is a polynomial over C, which is why they aren't a counterexample.

[marcinmozejko]

And x = 0 is not an open set.

[shoo]

agreed, x=0 is not an open set for n>1 dimensions, but it is "infinitely many points"