# Homework Help: How do I prove that a set is open?

1. Sep 27, 2010

### Raziel2701

1. The problem statement, all variables and given/known data
Let $$U = \{z \in {\mathbb C} | 0<|z|<1\}$$ Prove that U is an open set.

3. The attempt at a solution
I believe I have to use a tool known as the neighborhood or something:

$$\left|z - z_0\right|<\epsilon$$

Considering the boundaries, which just by looking at the set it seems to me that since they are not included then the set is open but how do I prove that? Anyway, looking at the boundaries, I'm thinking that I must go and pick some numbers such that the whole mess above with z -z_0 must be less than an epsilon, but that's as far as I go. How do I go about proving the set is open?

2. Sep 27, 2010

### vela

Staff Emeritus
For each point z in U, show there is an open ball centered at z that's contained in U.

3. Sep 27, 2010

### Raziel2701

This is the definition of open from my book:

A set $$A \subset {\mathbb C} = {\mathbb R}^2$$ is called open when, for each point $$z_0$$ in A, there is a real number $$\epsilon > 0$$ such that $$z \in A$$ whenever $$|z-z_0| < \epsilon$$

I have to satisfy this definition with the given set. But how?

4. Sep 27, 2010

### vela

Staff Emeritus
Just what I said. Show that for each point z0 in U, you can find some value of ε so that all points within ε of z0 are in U.

5. Sep 27, 2010

### vela

Staff Emeritus
To elaborate a bit, suppose z0=1/2; then ε=1/3 works because the circle of radius 1/3 centered at z0=1/2 completely fits in U. Note ε=1/4 would work as well. In fact, any ε up to 1/2 would work.

So the idea of the proof is to take some general point z0 in U and figure out what ε would work for that point. You'll probably find the triangle inequality useful.

6. Sep 27, 2010

### Raziel2701

I still don't know what to do.

So I have this picture in my head of what U is, namely all the points within the unit circle but not including its edge, nor the origin. So if I center z_0 so that it's between the origin and the edge of the unit circle, I can draw a disk whose radius would be epsilon, but then I figure I must somehow work with an epsilon that is a function of z_0. But I don't know how to do that.

That's as far as I go.

7. Sep 27, 2010

### vela

Staff Emeritus
Say $z_0 = re^{i\theta}$ where 0<r<1 and 0≤θ≤2π. Can you think of a value for ε in terms of r, θ, or both?

8. Sep 27, 2010

### Raziel2701

From your post I gather that epsilon would be r.

So I should be solving for r in this?

$$|z-re^{i\theta}| < r$$

Last edited: Sep 27, 2010
9. Sep 27, 2010

### vela

Staff Emeritus
It's not r. Consider the point z=0.9, for instance. If you drew a circle of radius 0.9 about the point z=0.9, it wouldn't lie within the unit circle.

10. Sep 27, 2010

### Raziel2701

Right, so there is a limit as to where exactly I must put z_naught. It can't be too close to the edge or else it encompass a part of the edge unit circle, and it can't encompass the origin either. I still am being way too obtuse to get this thing.

So I looked at a proof of a different set. And I thought that maybe I could write one similar to it.

I started like this:

Let z_naught be in U. Claim that D(z_naught; |z_naught|) is in U.

To see this, let z be in the disk D(z_naught;|z_naught|)..... then I guess |z-z_naught| must be equal to something else that will make it seem clear to me that I can use the triangle inequality to show that z_naught is in the disk, and thus the set is open??

I'm reaching a point of exhaustion here. It seems to me like the proof is essentially verifying the definition and making sure that the set U satisfies it, but I can't seem to understand how to even start the proof.

11. Sep 27, 2010

### vela

Staff Emeritus
No, you have to be able to find a neighborhood for every z0 in U. You do this by adjusting ε depending on how close z0 is to the edges.

You're not verifying the definition of open. You're applying it to U to show U is open.

A similar example would be to show the interval I=(0,1), a subset of the reals, is open. To show it's open, you want to find for every point x in I, there's an interval centered on x that's contained in I. The distance from x to the ends of the interval are x and 1-x. So choose ε=min{x, 1-x}/2. You should be able to see that the interval (x-ε, x+ε) is completely contained in I. Since you can do this for every point in I, the interval I satisfies the definition of an open set.

For the complex case, you want to do basically the same thing, but it's a little more work to show that the disc is contained in U. That's where the triangle inequality comes in.

12. Sep 28, 2010

### Raziel2701

0<|z-z0|<1

Is this what I must manipulate to show that |z-z0| is contained in U?

If that is the case, then the inequality I wrote up there should be equivalent to

|z-zo|^2 < (|x|+|y|)^2

right?

13. Sep 28, 2010

### Raziel2701

Actually wouldn't it have to be the reverse triangle inequality?

|z-z0| >= |x| - |y|

14. Sep 28, 2010

### vela

Staff Emeritus
Tell me what region the inequality $|z-z_0|<\varepsilon$ represents in the complex plane.

15. Sep 28, 2010

### Raziel2701

It's within the disk centered at $$z_0$$. The length $$|z-z_0|$$ must be less than epsilon, which I take to be the radius of the disk centered at z0.

I have to prove that this holds true for any value of z0 right?

16. Sep 28, 2010

### vela

Staff Emeritus
Yes, prove it's true given an appropriate choice for ε, so the first thing you need to do is figure out what value of ε would work if you're given z0.

It would probably help to draw the unit circle, an arbitrary point z0, and a circle of radius ε centered at z0 that lies within the unit circle. Hopefully, you'll see what limits how big ε can be for the given z0.