# Thread: Complex Analysis

1. ## Complex Analysis

My problem is as follows:

Let $A=\mathbb{C}$ ~ {z:Re z and Im z are rational}. Show that $A$ is a connected set.

I have begun my proof as follows:

Proof is by contradiction. Suppose that $A$ is disconnected. Then, by definition, there exist two disjoint open sets, $U$ and $V$, such that $A \subseteq U \cup V$, $A \cap U \neq \emptyset$ and $A \cap V \neq \emptyset$.

But then I need to come up with something that shows that the intersection of Re z and Im z exists and, therefore, $A$ 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.

2. ## Re: Complex Analysis

Originally Posted by tarheelborn
My problem is as follows:

Let $A=\mathbb{C}$ ~ {z:Re z and Im z are rational}. Show that $A$ is a connected set.

I have begun my proof as follows:

Proof is by contradiction. Suppose that $A$ is disconnected. Then, by definition, there exist two disjoint open sets, $U$ and $V$, such that $A \subseteq U \cup V$, $A \cap U \neq \emptyset$ and $A \cap V \neq \emptyset$.

But then I need to come up with something that shows that the intersection of Re z and Im z exists and, therefore, $A$ 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.
This isn't true. Note that $A$ is basically $\mathbb{Q}\times\mathbb{Q}$ and so if $A$ were connected then you would have that $\mathbb{Q}$ is connected since the projection map $\pi:\mathbb{Q}^2\to\mathbb{Q}$ is continuous and surjective. That said, $\mathbb{Q}$ is as disconnected as you could want (i.e. it is TOTALLY disconnected).

3. ## Re: Complex Analysis

But aren't Re z and Im z where both are rational two lines that intersect? Wouldn't that make them closed?

4. ## Re: Complex Analysis

Originally Posted by tarheelborn
But aren't Re z and Im z where both are rational two lines that intersect? Wouldn't that make them closed?
I'm not sure what you mean. The lines $\text{Re}(z)=q,\;\; q\in\mathbb{Q}$ is not open nor closed.

5. ## Re: Complex Analysis

The set A = the complex numbers minus {z: Re z and Im z are rational}. So A is, I believe, connected. I am just not sure how to get there...

6. ## Re: Complex Analysis

Originally Posted by tarheelborn
The set A = the complex numbers minus {z: Re z and Im z are rational}. So A is, I believe, connected. I am just not sure how to get there...
I think that D28 may well have misread the OP.
I also think that you need to simply show that $A$ is arcwise connected (path-connected) and therefore connected.

7. ## Re: Complex Analysis

Sorry, we haven't done path-connected and topology isn't a prerequisite for this class... I don't know how to do that. I am beginning to think they have the order of these classes VERY mixed up.

8. ## Re: Complex Analysis

Originally Posted by tarheelborn
Sorry, we haven't done path-connected and topology isn't a prerequisite for this class... I don't know how to do that. I am beginning to think they have the order of these classes VERY mixed up.
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 $(\alpha,t)$ is such that $\alpha$ is irrational then $\alpha+ti\in A$.
Then any point on the vertical line $\{\alpha+si:s\in\mathbb{R}\}$ is also in $A$.
Using those ideas of vertical and horizontal line segments you can construct a polygonal path between any two points in $A$ so that each point in the path is also in $A$. That proves that $A$ is connected.

9. ## Re: Complex Analysis

Thank you so much. I'll just have to work it out this way, then. I am pretty determined!

10. ## Re: Complex Analysis

I did indeed misread the OP. If you don't want to use path connectedness you could just note that $A$ can be written as a union of 'axes' which all intersect a single axis, and so is a union of intersecting connected sets.

11. ## Re: Complex Analysis

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.

12. ## Re: Complex Analysis

Originally Posted by tarheelborn
So I should abandon my proof by contradiction approach and go with a direct proof?
No there is no need to do that.
Lets go back to the OP.
Originally Posted by tarheelborn
Suppose that $A$ is disconnected. Then, by definition, there exist two disjoint open sets, $U$ and $V$, such that $A \subseteq U \cup V$, $A \cap U \neq \emptyset$ and $A \cap V \neq \emptyset$.
Suppose $z_1\in A\cap V~\&~z_2\in A\cap U$.
Is clear to you that the line segment from $z_1$ to $z_2$ must 'leave' $V$ and 'enter' $U$?
So some point on the segment not in $A$.
But if as I showed you there is a polygonal path from $z_1$ to $z_2$ which is a subset of $A$.
Those two ideas contradict one another.

13. ## Re: Complex Analysis

Now that makes sense. You're really good at making this clearer for me; it's not the first time you have helped me with a problem and I really appreciate it. Thank you.

14. ## Re: Complex Analysis

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 $A$ is disconnected. Then, by definition, there exist two disjoint sets, $U$ and $V$, such that $A\subseteq U \cup V$, $A \cap U \neq \emptyset$, and $A \cap V \neq \emptyset$. Now suppose that $z_1 \in A \cap V$ and $z_2 \in A \cap V$. Thus there is a line segment between $z_1$ and $z_2$ that lies partially in $V$ and partially in $U$ such that $[z_1,z_2]$ is connected by a finite union of segments of the form $[z_0,z_1]\cup[z_1,z_2]\cup \cdots \cup [z_{n-1},z_n]$, where the terminal point of one corresponds to the initial point of the next (when applicable). Thus, $[z_1,z_2]$ forms a polygonal path from $V$ to $U$. Hence $A$ is polygonally connected and, therefore, connected. $\square$

15. ## Re: Complex Analysis

Originally Posted by tarheelborn
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 $A$ is disconnected. Then, by definition, there exist two disjoint sets, $U$ and $V$, such that $A\subseteq U \cup V$, $A \cap U \neq \emptyset$, and $A \cap V \neq \emptyset$. Now suppose that $z_1 \in A \cap V$ and $z_2 \in A \cap V$. Thus there is a line segment between $z_1$ and $z_2$ that lies partially in $V$ and partially in $U$ such that $[z_1,z_2]$ is connected by a finite union of segments of the form $[z_0,z_1]\cup[z_1,z_2]\cup \cdots \cup [z_{n-1},z_n]$, where the terminal point of one corresponds to the initial point of the next (when applicable). Thus, $[z_1,z_2]$ forms a polygonal path from $V$ to $U$. Hence $A$ is polygonally connected and, therefore, connected. $\square$
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 $z_1,z_2\in A$ then you can connect them by a polygonal path consisting of just two intervals: a horizontal segment going from $z_1$ to a point with the same real part as $z_2$, followed by a vertical segment going from there to $z_2.$ Both segments lie entirely in A, and therefore A is polygonally connected.

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