# Open mapping theorem

August 21, 2012 Leave a comment

Today we’ll prove the open mapping theorem:

Theorem 1 (Open mapping Theorem)Let , for some open set in . Then , the set of all possible images of is either constant, or is open in .

Let , and let be a preimage. That is, . Since is open, there must be a small closed ball around completely contained in . Let be it’s radius. Then . Let . What do we know about ?

We know that the roots of are isolated points. After all, if not, we would have a convergent sequence of roots, contradiction one of the theorems we proved here. So we can choose to be even smaller so that the only root of in is . Since the boundary of is a circle, and hence compact, and is continuous it attains its minimum when restricted to the boundary. Let be the minimum value of for on the boundary of .

Now, by Rouché’s theorem, will have the same number of roots in in as for any in . (Application of Rouché’s theorem is left as an exercise.)

In particular, this means that for every complex number in , the function has at least one root in , and so is contained in , meaning our arbitrarily chosen point is in the interior of . Thus is open.