prove that the following subset of the complex plane is open. U=( x+iy : y>0)?

Answer 1

That depends on what you have as the basis for your open sets. You see, there is more than one definition of "open set" and there is more than one basis for those open sets. See:

http://en.wikipedia.org/wiki/Open_set
http://en.wikipedia.org/wiki/Topological_basis

Now, first note that:

U = ∪ { c + i y : y>0}

Now, I'm assuming you're using open balls as your basis. You can easily construct balls around the same intervals used in the union above and conclude that U = the union of those balls. Unions of open sets are always open, so U is open.

Or, if you want to show that it is open by showing it contains no point of its boundary:
http://en.wikipedia.org/wiki/Boundary_%28topology%29

This is simple:

Show the points along the line y = 0 (real line) are on the boundary. This is easy, since any open ball containing a real point x must have radius r>0, and then it contains a point in U and a point not in U, we could say x + i r/2 and its conjugate. You also have to show (trivially) that nothing else is on the boundary.

Answer 2

Your set U is the upper half-plane. This is open because given any point P in U there is a neighborhood of P (i.e., a circle centered at P) which is entirely contained in U. For example, if P is the point (x,y) with y > 0, the circle with center (x,y) and radius y/2 is in U.