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Topology of open set

  1. Feb 22, 2009 #1
    Topology of open set(newbie, I am stuck help!)

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

    Hi just found this found and have some basic questions about topology.

    If let say exist a metrix space (M,s) and two points [tex]x \neq y[/tex] in M. Then show that there exists open sets [tex]V_1,V_2 \in \mathcal{T}_s[/tex] such that [tex]x \in V_1[/tex] and [tex]x \in V_2[/tex] and finally [tex]V_1 \cap V_2 = \emptyset[/tex]

    2. Relevant equations



    3. The attempt at a solution

    I try to solve question 1)

    For V_1 to be an open set set then there must exist an open ball such that B(x)_r < r where is the radius of the ball which is M. Have I understood this correctly?
     

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    Last edited: Feb 22, 2009
  2. jcsd
  3. Feb 22, 2009 #2

    Dick

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    An open ball IS an open set. So why don't you just look for open balls around x and y which don't intersect? Use R=d(x,y), the distance between x and y to define them.
     
  4. Feb 22, 2009 #3
    Assuming s is the metric and Ts is the metric topology, a set V is in Ts if and only if for every element x in V, there exists some real number r such that the ball B(x, r) is contained in V (where B(x, r) is the set {a | s(x, a) < r}). The elements of Ts are called open sets in M under the topology induced by the metric s.
    First, make two balls, one around y and one around x. From the definition of metric, these balls at least contain one element each. Can you see where to go from there?
     
  5. Feb 22, 2009 #4
    Okay I will try to go on from there.

    what about question 3? I am told that I have use the fact

    [tex]d(x,z) \ leq d(x,y) + d(y,z) [/tex] and then show there does not exist a z which satisfies the above property of the metric and hence the intersection og V1 and V2 are the empty set. How does that sound?
     
  6. Feb 22, 2009 #5

    Dick

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    What are you doing? You haven't defined the radius of the open balls around x and y yet.
     
  7. Feb 22, 2009 #6
    Do you know the motivation for that inequality (The Triangle Inequality)? The triangle inequality is not necessary to show that there exists two disjoint open sets around unique elements in a metric space. What are the sets V1 and V2 that you are using? You have to have some type of definition before you can see whether their intersection is empty.
    Try letting V1 be B(x, r1) and V2 be B(y, r2). Picture this in the Euclidean plane. Pretend s is the Euclidean metric, the length of the straight line segment between x and y. What can you use s(x, y), the distance between x and y, for in order to define balls that don't intersect? After seeing how this works in the familiar setting of the Euclidean plane, make sure that your formula also makes sense in an arbitrary metric topology.
     
    Last edited: Feb 22, 2009
  8. Feb 22, 2009 #7

    Dick

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    Actually, you do want to use the triangle inequality to show the balls you chose don't intersect.
     
  9. Feb 23, 2009 #8
    The way I used to approach problems like this was to visualize them.. So the question is saying that I have some metric space, and the distance between points is given by some function d(a,b).

    I want to be able to (like your picture) draw a circle around arbitrary point x, and another one around arbitrary point y, so that the 2 circles don't intersect.

    So let me write down everything I know!
    x and y are 2 different points, so d(x,y) > 0. I don't know what the number is, but we can call it D.

    Now on a piece of paper that you pretend is your metric space, put a random point somewhere and label it x, put another point someone and label it y, then draw a line from x to y and call that line D.

    Can you draw a circle around x, and another circle around y that don't intersect?
    If you draw circles of the same size, how big can you make them before they intersect?

    Once you answer those questions, you should know what V_1 and V_2 are. How do you know they don't intersect? (besides looking at them!.. you don't!).. so assume there is a z in the intersection and see what that does to the triangle inequality.
     
  10. Feb 23, 2009 #9
    Hi all here is my trying to ansvar why

    [tex]V_1[/tex] is an open set of M and contains x.

    Let x' be a point in the ball B(x,r) and let s = ||x'-x||. We claim that the ball B(x'; r-s) is inside the ball B(x,r), if yes then its open. Then by taking the triangle inequality for any x in
    B(x'; r-s), then

    [tex]||x'-x|| = ||x-x' + x' -x| \leq \|x-x'\| + \|x'-x\| < r - s +s = r[/tex]

    and then if [tex]x \in V_1 \Rightarrow B(x,r) \subseteq M[/tex] for r> 0. Choose a y in B(u;r) is open by the above, then there exist r'>0 which implies that [tex]B(y;r') \subseteq B(x,r) \subseteq M[/tex] Then [tex]B(x;r) \subseteq V_1[/tex].
    Thus there every point in V1 is an inner point and V_1 is open.

    I repeat the same argument for V2. Would that be okay?

    Then in to show that [tex]V_1 \cap V_2 = \emptyset[/tex] that implies that doesn't exist an r > 0 such that [tex]B(x;r) \cap B(y;r) \neq \emptyset [/tex] in other words
    B(y;r') [tex]\notin B(x,r)[/tex] Thusly [tex]B(x;r) \cap B(y;r) \neq \emptyset [/tex] for all r. This means that neither x or y belongs to the intersection between V1 and V2.

    Finally I am required to construct a topology which does not meet any of the requirement from the above. Any hints or idears on howto do that?

    Cheers.
    Verystupidboyboy
     
    Last edited: Feb 23, 2009
  11. Feb 23, 2009 #10
    You have a lot wrong, do the picture thing like I told you.

    A general tip, as you're going along, make sure everything is defined, you can't say, let x' be a point in B(x,r).. what does that mean? (Try to answer my question, everything should be a learning experience because ultimately you cheat yourself if you skip things).

    My next hint to you (besides doing what I said before) is you need to construct 2 sets, V_1 and V_2 in such a way that V_1 contains x, V_2 contains y and they don't intersect each other. Just any old ball will not work.

    Ex.. let V_1 = B(x,r) then x is in V_1 since by def, V_1 contains all points less than r distance form the center (which happens to be x) and d (x,x) = 0 < r. You can do the same for V_2.. now you know that V_1 contains x and V_2 contains y, but these 2 balls can certainly overlap.. what's my mistake? I never defined r!

    Notice no matter how small r is, V_1 will contain x and V_2 will contain y, so just need them small enough that they don't intersect.
     
  12. Feb 23, 2009 #11
    Thank your answer,

    I thought I had got it now :frown: So my thought in the previous post about how to show that V1 and V2 are open is that completely wrong?
     
  13. Feb 23, 2009 #12
    I don't want to say too much to ruin this for you! That part is ok. Unless your professor specifically stated he wanted to see it, you don't need to show V_1 is open. By stating V_1 = B(x,r), you know its open since all balls in a metric space are open.

    You might want to prove it though just for practice.

    For me, I don't understand what you mean when you write:
    if x in V_1 -> B(x,r) is a subset of M for r > 0.

    Everything is a subset of M!

    First, create V_1 (somehow, this is the fun part)
    Second, show V_1 contains x.
    Third, do the same for V_2 and y.
    Fourth, show their intersection is empty.

    You may want to show that V_1 and V_2 are open, but I don't think that's the point of this assignment.

    also, why is | x' - x | = | x - x' + x' - x| because | x - x' + x' - x| = | x - x + x' - x'| = |0|!
     
    Last edited: Feb 23, 2009
  14. Feb 24, 2009 #13
    How do I then show that there exist a topology space which does not comply with either of the above statements?
     
    Last edited: Feb 24, 2009
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