Kouznetsov Wrote:bo198214 Wrote:...Why do you think that this set is open?

And that implies that , which contains an open non-empty set, always contains points such that .

...

Just take a point of the interior of such that (if such a point does not exist then is constant), then there is a neighborhood of such that is bijective on . That means and are continuous (because analytic). And the preimage of an open set is open under a continuous map. So the preimage of under is open and so we find arbitrary many points in such that .

Kouznetsov Wrote:Do you make any difference between and ?

Oh no its both the same here.