Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Sep 27, 2010 #1
    1. The problem statement, all variables and given/known data
    Let [tex]U = \{z \in {\mathbb C} | 0<|z|<1\}[/tex] 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:

    [tex]\left|z - z_0\right|<\epsilon[/tex]

    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. jcsd
  3. Sep 27, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    For each point z in U, show there is an open ball centered at z that's contained in U.
     
  4. Sep 27, 2010 #3
    This is the definition of open from my book:

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

    I have to satisfy this definition with the given set. But how?
     
  5. Sep 27, 2010 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  6. Sep 27, 2010 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  7. Sep 27, 2010 #6
    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.
     
  8. Sep 27, 2010 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Say [itex]z_0 = re^{i\theta}[/itex] where 0<r<1 and 0≤θ≤2π. Can you think of a value for ε in terms of r, θ, or both?
     
  9. Sep 27, 2010 #8
    From your post I gather that epsilon would be r.

    So I should be solving for r in this?

    [tex]|z-re^{i\theta}| < r[/tex]
     
    Last edited: Sep 27, 2010
  10. Sep 27, 2010 #9

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  11. Sep 27, 2010 #10
    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.
     
  12. Sep 27, 2010 #11

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
  13. Sep 28, 2010 #12
    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?
     
  14. Sep 28, 2010 #13
    Actually wouldn't it have to be the reverse triangle inequality?

    |z-z0| >= |x| - |y|
     
  15. Sep 28, 2010 #14

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Tell me what region the inequality [itex]|z-z_0|<\varepsilon[/itex] represents in the complex plane.
     
  16. Sep 28, 2010 #15
    It's within the disk centered at [tex]z_0[/tex]. The length [tex]|z-z_0|[/tex] 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?
     
  17. Sep 28, 2010 #16

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook