# Complex analysis planar sets

1. Oct 23, 2011

### nickolas2730

I dont get the meaning of "connected" in the chapter of planar sets.
The text book said " An open set S is said to be connected if every pair of points z1, z2 in S can be joined by a polygonal path that lies entirely in S"

So do i just randomly pick 2 points in S to check if they are both in S?
but if i pick a point in S, how can it not being in S....
and my questions is, what points do i pick to check if it is connected.

Thanks

2. Oct 23, 2011

### lavinia

you need to check that the path connecting them can be chosen to lie completely in S.

3. Oct 23, 2011

### HallsofIvy

You cannot just pick two points. In order to show that a set is "connected" in this sense, you would have to show that any two points have such a have such a path connecting them. And you do not just check that two points "are in S", you show that there exist a polygonal path connecting them that lies in S- that is, all points on the path are in S.

Two prove that a set is NOT connected you only need to give a "counter-example" - two points in the set such that a path connecting them cannot lie in the set. For example the set {(x, y)| xy> 0}, where x and y are numbers, is not connected since any path from (1, 1) to (-1, -1) must pass through a point where either x or y is 0- and such a point is not in the set.

4. Oct 23, 2011

### nickolas2730

Thank you so much,
as what i get, it is super hard to be connected except the set is very specific, right?
like, it makes many constrain for x and y.

5. Oct 23, 2011

### lavinia

Connected set are easy. Any open ball is connected. The entire plane is connected. Actually your definition is of a path connected set. It is possible for a set to be connected but not path connected.

6. Oct 23, 2011

### mathwonk

but not for an open set in euclidean space. there the concepts coincide.