My problem is as follows:
Let ~ {z:Re z and Im z are rational}. Show that is a connected set.
I have begun my proof as follows:
Proof is by contradiction. Suppose that is disconnected. Then, by definition, there exist two disjoint open sets, and , such that , and .
But then I need to come up with something that shows that the intersection of Re z and Im z exists and, therefore, is connected, but I am not sure how to say it. I'm not exactly sure I am on the right track, but it seems like this should work. I will appreciate your help. Thanks.
Sorry to say, I don't think this can be done with separated set alone.
In a plane any pathwise connected set is connected.
Here is the idea. Suppose that is such that is irrational then .
Then any point on the vertical line is also in .
Using those ideas of vertical and horizontal line segments you can construct a polygonal path between any two points in so that each point in the path is also in . That proves that is connected.
So I should abandon my proof by contradiction approach and go with a direct proof? Hmmm.... Ok, I can try it from that direction, although I was seeing A more as the union of space that eliminated the two lines Re z and Im z, but I will be the first to admit I have a world of trouble visualing these complex problems.
No there is no need to do that.
Lets go back to the OP.
Suppose .
Is clear to you that the line segment from to must 'leave' and 'enter' ?
So some point on the segment not in .
But if as I showed you there is a polygonal path from to which is a subset of .
Those two ideas contradict one another.
To clarify that I am on the right track in the formal proof, may I run the final version by you?
Proof. (by contradiction). Suppose that is disconnected. Then, by definition, there exist two disjoint sets, and , such that , , and . Now suppose that and . Thus there is a line segment between and that lies partially in and partially in such that is connected by a finite union of segments of the form , where the terminal point of one corresponds to the initial point of the next (when applicable). Thus, forms a polygonal path from to . Hence is polygonally connected and, therefore, connected.
I think that you were unnecessarily deterred by the use of the terms path-connected and arcwise connected in an earlier comment. It seems that you know what is meant by the term polygonally connected, and that polygonally connected implies connected. What the earlier hints were saying is that if then you can connect them by a polygonal path consisting of just two intervals: a horizontal segment going from to a point with the same real part as , followed by a vertical segment going from there to Both segments lie entirely in A, and therefore A is polygonally connected.