How do I prove that a set is open?

• Raziel2701
In summary, to prove that the set U = {z ∈ C | 0 < |z| < 1} is open, we must show that for every point z0 in U, there exists an open ball centered at z0 that is contained in U. This can be done by choosing an appropriate value for ε, the radius of the open ball, which would depend on the distance of z0 from the boundaries of U. By using the triangle inequality, we can show that for any z0 in U, there exists an ε that satisfies the definition of an open set. Therefore, U is an open set.
Raziel2701

Homework Statement

Let $$U = \{z \in {\mathbb C} | 0<|z|<1\}$$ Prove that U is an open set.

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?

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

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?

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.

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.

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.

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?

From your post I gather that epsilon would be r.

So I should be solving for r in this?

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

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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.

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.

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.

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?

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

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

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

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?

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.

What is an open set?

An open set is a set of points in a topological space where every point has a neighborhood contained entirely within the set.

How do I prove that a set is open?

To prove that a set is open, you can use the definition of an open set, which states that every point in the set must have a neighborhood contained entirely within the set. You can also use the concept of interior points, where every point in the set is an interior point.

What are some common techniques for proving a set is open?

Some common techniques for proving a set is open include using the definition of an open set, showing that every point in the set is an interior point, and using the concept of continuity to show that a function maps open sets to open sets.

Can a set be both open and closed?

Yes, in certain topological spaces, a set can be both open and closed. This is known as a clopen set. An example of this is the empty set and the entire space in the discrete topology.

What is the difference between an open set and a closed set?

An open set is a set where every point has a neighborhood contained entirely within the set, while a closed set is a set that contains all of its limit points. In other words, an open set does not contain its boundary points, while a closed set does.

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