Via topology, here's what I think is the natural extension of the intermediate value theorem to higher dimensions: Let U be a nonempty open set in R^n. Let f: U -> R^n be continuous. Suppose there is a closed topological n-ball B in U, and a point c in R^n, such that 1) f restricted to the boundary of B -- a topological (n-1)-sphere that we call S -- never takes the value c 2) The mapping K_f: S -> S^(n-1) (the standard unit (n-1)-sphere) defined by (K_f)(x) := (f(x) - c)/ ||f(x) - c|| is not homotopic to a constant map (i.e., has degree unequal to 0). --------------- Theorem: Then f, restricted to B, takes the value c in the interior int(B) of B. --Dan On 2013-02-12, at 6:18 PM, rcs@xmission.com wrote:

The two-dimensional version of the intermediate value theorem.

Suppose F(z) is a continuous function from C->C over the complex numbers, or at least over the unit square. We are silent about the differentiability of F. Suppose that the real part of F, on the vertical edges of the square, has ReF(0+iy)=0 and ReF(1+iy)=0, for 0<=y<=1. On the horizontal edges, the imaginary part of F is similarly constrained: ImF(x+0i)=0 and ImF(x+1i)=0, for 0<=x<=1. (This forces F(z)=0+0i at the four corners of the square.)

I presume that this implies that there is some point z inside the square with F(z) = 0+0i.

The puzzle is to come up with a proof, ideally adaptable to a higher number of dimensions.

The one-dimensional version can be proved by (iterated) bisection of the domain, with no fancy topology needed. This doesn't seem to extend to higher dimensions though.