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General topology: Prove a Set is Open

  1. Sep 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Let A:={x∈ℝ2 : 1<x2+y2<2}. Is A open, closed or neither? Prove.


    2. Relevant equations
    triangle inequality d(x,y)≤d(x,z)+d(z,y)

    3. The attempt at a solution
    First I draw a picture with Wolfram Alpha. My intuition is that the set is open.
    rh5mwj.png
    Let (a,b)∈A arbitrarily and r=min{√(a2+b2)-1, √2 -√(a2+b2)} (by geometry). It is clear that then Br(a,b)⊂A but it needs to be formally proven. Let z∈Br(a,b) and now I need to show that ||x-z||=d(x,z)<r.
    d(x,z)≤....I am suck here.

    And the next thing would be to prove the set is not closed. Its complement Ac={x∈ℝ2 : x2+y2≤1}∪{x∈ℝ2 : x2+y2≥√2} is clearly closed, though.

     
    Last edited: Sep 10, 2016
  2. jcsd
  3. Sep 10, 2016 #2

    andrewkirk

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    To prove the set is open you need to show it can be constructed from given open sets using the allowed operations, which are arbitrary unions and finite intersections.

    What have you been given as the original set of open sets for the topology of ##\mathbb R^2## (known as the 'basis')?

    There are two common bases used for ##\mathbb R^2##.. One is the set of all open boxes of form ##(a,b)\times(c,d)## for different combinations ##a,b,c,d## and the other is the set of all open balls of form ##\{(x,y)\ :\ \|(x,y)-(x_0,y_0)\|<r\}## for different combinations of ##x_0,y_0,r##. The approach to proving openness will depend which of these bases you are using (or other basis, if you have been given that).
     
  4. Sep 10, 2016 #3
    I am not familiar with that approach. We are using open balls.
     
  5. Sep 10, 2016 #4

    Krylov

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    Edited, no longer interested in contributing to this thread.
     
    Last edited: Sep 10, 2016
  6. Sep 10, 2016 #5

    andrewkirk

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    It's easy using balls. Your set is the intersection of two open sets. One is an open ball and the other is the complement of a closed ball.
     
  7. Sep 10, 2016 #6
    May be easy but how to prove it formally?
     
  8. Sep 10, 2016 #7

    PeroK

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    If you can assume that an open ball is open and that the intersection of two open sets is open, then the proof should be easy. But, showing that an open ball is open would seem to me to be not much different from showing that the set you have is open.

    So, perhaps your original approach is the one that is expected: show that each point in A has a neighbourhood in A. If you continue with this approach, you might consider how far points are from the origin.
     
  9. Sep 10, 2016 #8
    Actually we have proven in class that open ball is open set and that a finite intersection of open sets is open.

    Now A=B((0,0),r1)∩B((0,0),r2)={y∈ℝ2: d(y,(0,0))≤1)}c∩{z∈ℝ2: d(z,(0,0))<√2}

    {y∈ℝ2: d(y,(0,0))≤1)} is closed ball by definition and therefore its complement {y∈ℝ2: d(y,(0,0))≤1)}c is open? And intersection of two open sets is open so A is open.

    I think this approach is more elementary.
     
    Last edited: Sep 10, 2016
  10. Sep 10, 2016 #9

    andrewkirk

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    That is not consistent with your response in post 3 to my question about what basis (or sometimes written as 'base') you have been given. If you are given the open balls as a basis then there it is redundant to prove the openness of an open ball. It is open by definition of the topology.

    Also, it is redundant to prove that a finite intersection of open sets is open. Such an intersection is open by definition.

    Any topology must be given either by specifying a set of open sets - which are simply specified, not proven - or it is constructed from other topologies, as in product or box topologies. I'm guessing you haven't dealt with the latter yet, so you must have been given a specified collection of open sets to start with. What is that collection?
     
  11. Sep 10, 2016 #10
    Now I am even more confused. :eek:
     
  12. Sep 10, 2016 #11

    micromass

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    andrew, the OP is likely only acquainted with metric spaces and not with any topology. So the word topology or bases are likely foreign to him.
     
  13. Sep 10, 2016 #12
    so, how to proceed?
     
  14. Sep 10, 2016 #13

    PeroK

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    You need to start the way you started in post #1:

    And now, think about the distance of ##z## from the origin. Hint: use your relevant equation!
     
  15. Sep 10, 2016 #14
    x=(x1, x2) ∈A
    and z=(z1,z2)∈B(x,r)

    d(x,z)=√((x12-z12)+(x22-z22))<r.

    d(z,(0,0))=√(z12+z22)≤√((x12-z12)+(x22-z22)) + √(x12+x22)<r+√(x12+x22)<...?

    (triangle inequality)
     
  16. Sep 10, 2016 #15

    PeroK

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    It might be simpler to work the metric notation throughout. You could even rewrite the definition of ##A##:

    ##A = \{x: 1 < d(0, x) < \sqrt{2} \}##

    This is good, because it means you are using the properties of any metric, not just the usual metric for ##\mathbb{R}^2##

    And you have ##r = min\{d(0, x) - 1, \sqrt{2} - d(0, x) \}##

    Now, you had: Let ##a \in A## ...

    Can you pick it up from there?
     
  17. Sep 10, 2016 #16
    d(a,z)<r.

    d(0,z)≤d(0,a)+d(a,z)<r + d(a,0)

    i am confused and topology is giving me brain cancer soon
     
  18. Sep 10, 2016 #17

    PeroK

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    Well, okay, but what next?

    Also, you need to pay attention. My last post wasn't exactly perfect, so take a look and fix it!
     
  19. Sep 10, 2016 #18
    d(a,z)<r.

    d(0,z)≤d(0,a)+d(a,z)<r + d(a,0)<√2?
     
  20. Sep 10, 2016 #19

    PeroK

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    Why is that?
     
  21. Sep 10, 2016 #20
    That's what I need to show next somehow :oldconfused:
     
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