# Connectedness of a Given Set in the Complex Plane

1. May 11, 2014

### NihilTico

1. The problem statement, all variables and given/known data

Let $A=\mathbf{C}-${$z:Re(z)$ and $Im(z)$ are rational}. Show that $A$ is a connected set.

2. Relevant equations

My book gives the definition of a disconnected set as a set that satisfies three conditions. A set $A$ is disconnected if there exist two open sets $U$ and $V$ in $\mathbf{C}$, that satisfy (i) $A\cap U\cap V=\emptyset$; (ii) $A\cap U\ne\emptyset$ and $A\cap V\ne\emptyset$; (iii) $A\subset U\cup V$.

3. The attempt at a solution

I feel that I should assume that the set is disconnected and derive a contradiction from this, but I haven't an idea where to start when I do that. Another approach I entertained was to show that there exists a polygonal arc between any two points such that it is entirely contained within the set $A$, of course, I couldn't figure out how that would work either.

So, going off of the former attack mentioned, assuming that the set $A$ is disconnected only gives me that there exist two sets that satisfy the above three arguments. My intuition in this case tells me that a contradiction would most easily arise by considering (i) and (iii). But isn't it the case that I could construct two sets that satisfy all three of these conditions by requiring $U$ to be the set of all pairs (x,y) such that at least one of x and y are rational and by letting $V$ be the set of all pairs (x,y) such that both of x and y are irrational? I see no contradiction in that, and it seems to say to me that it is disconnected.

Last edited: May 12, 2014
2. May 11, 2014

### Dick

The U and V you have picked aren't open.

3. May 11, 2014

### NihilTico

That's true. But if I were use this tack, I still end up at a dead end.

Should I let a function define $A$. For instance, $f(z)$ that is defined only when $z\in A$?

4. May 12, 2014

### Dick

No, concentrate on defining a path connecting any two points in A. It's not as hard as you might think. For example if a line has irrational slope than only one point with rational coordinates can lie on it.

5. May 12, 2014

### pasmith

The set A consists of those $z \in \mathbb{C}$ such that at least one of $\mathrm{Re}(z)$ and $\mathrm{Im}(z)$ is irrational.

Can you show that if $z \in A$ then at least one of the horizontal line $\{w \in \mathbb{C} : \mathrm{Im}(w) = \mathrm{Im}(z)\}$ and the vertical line $\{w \in \mathbb{C} : \mathrm{Re}(w) = \mathrm{Re}(z)\}$ lie wholly within $A$?

6. May 12, 2014

### NihilTico

I'm not sure I understand, because it would seem to me that there exists no rational number, if $\pi$ and $\imath$ represent two irrational numbers, such that $\pi\cdot\frac{p}{q}+\imath=\frac{a}{b}$. I can't figure out how to show that exactly one rational number exists such that $\pi{p}+\imath{q}$ is rational.

But other than that, I think I see what you are getting at, if I follow correctly, what you are saying is that I can break $A$ up into open and connected subsets such that each segment component of of $A$ contains one point in common, say $z_0$ which would mean that their union is a connected set. The proof that the union of connected subsets of some set, say B, that contain some given point in common is connected is fairly straightforward as we can demonstrate that, since each member of the union is connected and contains one point in common, each member of B must be a subset of only one set.
So if I show that a general segment, $I$, in the complex plane with an initial point of $z_0$ and some end point of $z$ satisfies the condition that it not be defined at any point where it has rational real and imaginary parts, then $\bigcup\limits_{z\ne{z_0}}I_z$ is the union of open and connected sets where each member of the union contains the point $z_0$, so it is both open and connected, as I would have shown that the set $A$ lies in one set alone, and not in the union of two disconnected sets or something like that?

So the only part I'm hung up about is that exactly one rational number exists so that a line is defined there and has a rational value. I think that it works if the line is defined through the origin, for example $f(x)=\pi\cdot{x}$ but otherwise I'm struggling to see it.

7. May 12, 2014

### NihilTico

If I understand what you are saying, if $z\in{A}$ then $Re(z)\in\mathbf{Q}$ or $Im(z)\in\mathbf{Q}$. So there are three cases: (i) $Re(z)\in\mathbf{Q}$ and $Im(z)\notin\mathbf{Q}$ (ii) $Im(z)\in\mathbf{Q}$ and $Re(z)\notin\mathbf{Q}$ and (iii) both $Re(z)\in\mathbf{Q}$ and $Im(z)\in\mathbf{Q}$. If (i) is true, then the horizontal line defined by $Im(z)$ lies wholly within $A$, if (ii) is true, then the vertical line defined by $Re(z)$ lies wholly within $A$ and the obviously if (iii) is true, then both the horizontal and vertical line lie in $A$.

8. May 12, 2014

### Dick

You are making it too complicated. Suppose a line contains two rational points. What can you tell me about its slope?

Now you know lots of lines that contain at most one rational point and a lot of horizontal and vertical lines that contain no rational points. Splice them together to make a path between two points in A. You might want to make some cases. Suppose between the two points you have an irrational x coordinate and an irrational y coordinate. Then what?

Last edited: May 12, 2014
9. May 12, 2014

### NihilTico

1. If a line contains two rational points, then its slope is rational: $\frac{y_1-y_2}{x_1-x_2}$.
2. If, between the two points there are two irrational points, then I would have to find a polygonal arc contained only in $A$ that intersects that point?

I'm not sure where you are headed with this now. The book I'm using proves that the union of a collection of connected sets sharing one point in common is connected. So would it be that, if I can show that any point that has at least on irrational coordinate can be connected by an open path contained in $A$, that I will have shown that $A$ is connected?

So maybe what you're getting at here is that it would be something along the lines of saying that the path between $z_0=(p,q)$ where p and q are irrational and $z=(a,b)$, where at least one of a and b is irrational, is contained solely in $A$ as for any point $z$ where at least one of $a$ and $b$ is irrational, that there exists a vertical and/or horizontal line in the complex plane through this point that is also contained within $A$ such that it intersects a line with an irrational slope that is contained within $A$ as well. Meaning that we will have constructed a path through $A$ to that point?

And bouncing off the assumption that a line with an irrational slope contains at least one rational pair of coordinates, if it is the case that there is a rational pair between $z_0$ and the horizontal and/or vertical line that I am seeking to intersect, that we can change the slope of the path through the complex plane as to avoid that point and still intersect that line?

10. May 12, 2014

### Dick

In your sample case you have a horizontal AND a vertical line through $z_0$ that contain no rational points. And you have either a horizontal or a vertical line through $z$ that contain no rational points. Two of those lines must intersect. Describe a path through A that connects $z_0$ and $z$. You don't even need any irrational slope lines in this case. In what kind of case can't you rely on those particular horizontals and verticals?

Last edited: May 12, 2014
11. May 12, 2014

### NihilTico

So, given a point $z_0=(p,q)$ of $A$ where $p,q\in\mathbf{C}-\mathbf{Q}$, the path between $z_0=(p,q)$ and $z=(a,b)$ where at least one of $a,b$ are irrational is given as the union of the line segments.

If $a$ is irrational, then this path is given by the union of $I_1\cup{I_2}$ where $I_1$ is defined by the function $f(t)=p\cdot{t}+i\cdot{q}$ for $t\in[0,\frac{a}{p}]$ and $I_2$ is given by $g(t)=a+i\cdot{q}t$ for $t\in[0,\frac{b}{q}]$.

If $b$ is irrational, then this path is given by the union of $I_1\cup{I_2}$ where $I_1$ is defined by the function $f(t)=p+i\cdot{q}t$ for $t\in[0,\frac{b}{q}]$ and $I_2$ is given by $g(t)=p\cdot{t}+i\cdot{b}$ for $t\in[0,\frac{a}{p}]$.

If both $a$ and $b$ are irrational, either of the paths taken before are sufficient.

Does that work? Is it sufficient to demonstrate that the set is connected?

So I presume you mean that I can't rely on the both vertical and horizontal if I chose an initial point $z_0=(p,q)$ where exactly one of $p$ and $q$ were irrational?

12. May 12, 2014

### Dick

I think you are thinking about it correctly, but your exact parametrizations are wrong. Just draw the picture first. Take the a irrational case, then you want to connect a+ib to a+iq and then connect that to p+iq. And the tougher case is say, b and q both rational. Then your lines through both points are both vertical. They don't intersect. Then you can either play the irrational slope game, or you could introduce a third point in A that does what you need. Oh, and p and q aren't elements of C-Q. They are elements of R-Q. TeX it up later. Just try to get the idea first.

13. May 12, 2014

### NihilTico

Alright. As I understand it, the idea is that I want to show, given some point in z_0 in A, that I can produce a path in A to any other point in A. Right? So I think that it should suffice if I consider the point z_0 to have at least one coordinate as irrational and to have my end point have at least one coordinate irrational as well and then I should be able to easily show the cases where z and z_0 have pairs of irrational coordinates. I'm not sure how I would demonstrate that the path taken along an irrational slope would not run into a pair of rational coordinates, and that, if it did, how I would show that I could append another part to the path to avoid those coordinates.

Oh, and I was letting z_0 be the starting point, that was a little unclear, my apologies!

14. May 13, 2014

### Dick

Hmm. I don't seem to be communicating well. As I asked, if a line has irrational slope can it contain two points with rational coordinates? You should say no for a reason you've already given. You can use that fact if you want - but you can also just use the horizontals and verticals. How do you get from p+iq to a+ib where q and b are rational? Hint: introduce a third point that has a path to each.

15. May 13, 2014

### NihilTico

When you say two points with rational coordinates, you do mean something like point A with rational coordinates a+ia' and then a second point B with coordinates b+ib', correct?

So the point p+iq and a+ib have vertical 'strips' that lie entirely in the set A, which means that for all values of, say c, in p+ic, p+ic lies within that strip and c is irrational, since c can be any real number. So that is our third point, p+ic. So that means there is a horizontal 'strip' at c which must intersect the vertical strip located at the coordinate a+ic and then I simply take a path down or up from there to b. Is my reasoning sound there?

Last edited: May 13, 2014
16. May 13, 2014

### Dick

Yes, that's what I mean by rational coordinates, and yes, your reasoning is sound. It's actually much easier to understand without the parametrizations, isn't it?

17. May 13, 2014

### NihilTico

Yes, it is much easier to look at it generally, without the parameterizations (less room for error!), but to actually prove it, won't I need to explicitly define a path through parameterizations? Is it sufficient to show, for a general starting point in the A, that I can connect any two points in A?

Last edited: May 13, 2014
18. May 13, 2014

### Dick

Whether you need to actually show the parametrizations if you have a good description of the path would be up to your instructor. If you take a point p where both coordinates are irrational, that can connect to any other point of A, right? To connect any two points of A, you could just take a path through p.

19. May 13, 2014

### NihilTico

I see. I don't actually have an instructor, so, in spirit of being as explicit as possible, I might want to parameterize the path taken to some arbitrary point, but, otherwise, it is perfectly acceptable to go about it without parameterization?

20. May 13, 2014

### Dick

For me it is. If you say vertically from a+bi to a+qi then horizontally to p+qi that describes the path perfectly without me needing to see the t's.