Can some one please help me with this, why is third set in the picture not simply connected. The definition of simply connected (in space of complex numbers) is:
A set is said to be simply connected if every closed path in the set is homotopic to a point.
so does this definition mean it must be homotopic to all points in our set or must there exist a point. Cause if it says there must exist a point then the third set must be simply connected right?
On
A good way to describe the third picture would be to say it has simply connected (path-)components. In a sense it's almost as good as simple-connectedness since the common things you would need simply-connected open sets for (integrals of holomorphic functions around closed curves equal to 0, an antiderivative for every holomorphic function, etc.) work just as well by considering the components separately. But as a whole it is not simply connected because it has more than one connected component.
A path-connected topological space $X$ is simply connected if for any given point $x_0$ on $X$, any loop $\sigma$ based at $x_0$ is path-homotopic to the constant loop $e_{x_0}$ at $x_0$.
The third space (subset of $\Bbb C$) is not connected, so there is no path between a point in the first component and a point in the other, say. Thus this space is not path connected, hence simple connctedness is not defined.