I'm studying introductory complex analysis right now with Marsden's Basic Complex Analysis, and I'm stuck on understanding why the following are connected and/or compact (it's problem 1.4.15 on Marsden; the book has the solutions, but I don't really understand why there are what they are):
1) {$z$ | $1 \le |z| \le 2$} - Connected and Compact
Is this because the set is closed and bounded by $1$ and $3$? I also have no clue how to apply the "not connected" definition.. I need to show that this set is not a subset of some $2$ open sets, right? How do you generally go about doing that?
2) {$z$ such that $|z| \le 3$ and |Re $z$| $\ge 1$} - Compact, Not Connected
I don't quite see how this set is closed and bounded.. is it simply $1$ and $3$ as well?
3) {$z$ such that |Re $z$| $\le 1$} - Connected, Not Compact
4) {$z$ such that |Re $z$| $\ge 1$} - Neither
Completely lost on these.
Any help with me understanding why the answers are what they are would be extremely helpful. Thank you.
•By definition, if it can be written as the disjoint union of open (hence also closed; hence clopen) sets, the region isn't connected...
•Closed and bounded is equivalent to compact (by Heine-Borel)..
For (1), the region is an annulus( the region between two circles) ; hence compact and connected (it's all of one piece, and closed, since it includes its boundary, and bounded)
(2): The region is two disjoint subsets of the closed disk of radius $3$. It's not connected, since the union of these two regions, which are clopen, is the whole space. . Easily seen to be closed and bounded...
(3): This is a an infinite vertical strip of width $2$. Clearly connected (one piece, intuitively) , since unbounded, not compact...
(4): This is the region outside the strip in (3), but including the boundary. Clearly unbounded and with a big "hole" in it, it is the union of two clopen disjoint regions. So neither..