Math Help - Complex analysis - connected sets

1. Complex analysis - connected sets

Note: Sorry if this is in the wrong section, I wasn't sure where to put it.

The question
Are the following regions in the plane 1) open 2) connected, 3) domains?
a) the complement of the unit circle
b) the real numbers

My attempt
a) yes, yes, no
However the solution is yes, no, no. I don't understand why this isn't considered connected. The way I see it, the compliment of the unit circle is everything outside the circle. So there's always a 'path' from one element in the set to another, if I'm not mistaken. Am I missing a detail here?

b) yes, yes, yes
The correct solution is no, yes, no. Why are the real numbers considered a closed set? I'm a little confused, since real numbers can be 'as big as you like'. I think I'm missing some intuition here. I'm also not sure why it isn't considered a domain, but I guess that's because I believe it's open. I realise it can't be a domain if it's not both open and connected.

Any assistance would be greatly appreciated!

2. If you're not sure where to put it, report your own post and a mod will move it for you.

a). My guess is that your book defined "connected" as "simply connected", which is a bit odd. Either that, or it's a typo (or maybe a "thought-o"). I would have thought your answer correct, but I could definitely be wrong.

[EDIT]: See Plato and HallsofIvy's posts below for the correct explanation.

b). The real numbers are being considered as a subset of the complex plane, not just the real numbers like in real analysis. So, in looking at the neighborhoods around points in the real line, those neighborhoods are going to be two-dimensional. And you can see that any such neighborhood around a single point on the real line will have points in it that are not in the real line. So the real line is still closed, but it is no longer, as it is in real analysis, open. You see how that works? Essentially, the definition of "open set" has now changed, because your definition of "neighborhood" has changed.

3. Thanks Ackbeet. Their idea of 'connected' is indeed confusing. In our lecture today, the lecturer demonstrated it by drawing a line from one point in the set to another point, so that's the idea in my head.

4. The line segment that your lecturer drew: did he say anything about the entire line segment remaining in the set? As in, kind of a convex idea? Because, if all such line segments drawn have to stay inside the set (which you can easily see is not true for the complement of the unit circle), then you're really talking about simply connected sets.

5. I'll quote his notes:

"S is connected if any two points can be joined by a continuous curve lying entirely in S. (Intuitively, we think of S being in one piece.)"

6. The complement of the unit circle is open but not connected.
Recall that the unit circle is just $\{z:|z|=1\}$ so connecting a point in the interior to a point in the exterior must cross the circle.

The complement of the reals is just the upper half-plane union lower half-plane. SO?

7. That definition is, technically, "path connected" which is not exactly the same as "connected". One can show that all path connected sets are connected but there exist connected sets that are NOT path connected.

In any case, I think your problem is confusion between "circle" and "disk". The problem says "circle" and you (and Ackbeet apparently) are thinking "disk". The unit circle in the complex plane is the set of points, z, satisfying $z^2= 1$, while the unit disk is the set of points, z, satisfying $z^2\le 1$. See the difference?

The point z= 0 is inside the unit circle, the point z= 2 is outside it but both are in the complement of the unit circle. Since they cannot be connected by a path that does NOT cross the unit circle, the complement of the unit circle is not (path) connected.

Blast! Once again I typed too slow and Plato beat me!

8. Reply to Plato and HallsofIvy:

Good catch on the unit circle. My bad.