Let be the set of all points in which can be polygonally connected and be the set of all points in which cannot be polygonally connected. Prove that and are open sets. Now since and it follows that or by definition of connectdness of .3 I think is partially done for me in my notes, I'll see if I can do it
but hints are appreciated
Note: This argument can be used to show that there is a polygonal path consisting of only vertical and horizontal segments.
If there is with then is twice differenciable too (in fact it is infinitely differenciable) and this would mean that the mapping is differenciable - but that is false - it is not differenciable at .4 I used the unit circle at (1,1) showed the integral wasn't 0 and then said the function wasn't analytical since the integral wasn't 0 on a closed curve <- can someone confirm this is right?