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Homework Help: Topology- Hyperplane proof don't understand

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Let [tex] \mathbf {a} \in R^n [/tex] be a non zero vector, and define {[tex] S = \mathbf {x} \in R^n : \mathbf {a} \cdot \mathbf {x} = 0 [/tex] }. Prove that S interior [tex] = {\o} [/tex]

    2. Relevant equations

    3. The attempt at a solution

    Intuitively I understand that if a is a vector in R^3, S would be a plane. And if I place an open ball on any point on the plane it would include points that are not on the plane. So S interior would be empty.

    But I'm not sure how to prove this rigorously?
    I'm supposed to somehow use the hint that a hyperplane in R^n will have 1 dimension less than R^n (like the plane in R^3 has 2 dimensions)
    But I am confused about the dimension of the plane? All vectors on a plane in R^3 will have 1 (the same) component = 0? So does this mean that the dimension of the plane is 2?
  2. jcsd
  3. Sep 29, 2010 #2
    Can I just show that there is a point in the ball that has n components ?
    Instead of n-1
  4. Sep 29, 2010 #3


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    If a.x=0 you can pretty easily find a point y in any ball around x such that a.y is not zero.
  5. Sep 30, 2010 #4
    I'm a little confused about that.
    Do I need to differentiate between a point and a vector?
    Like if I show that there are points in the ball that are not on the plane, then do I need to specify that the vector corresponding to that point is not perp a ?
  6. Sep 30, 2010 #5


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    Now I'm confused about that. Any point y in the ball around x of radius r is given by y=x+a where 'a' is any vector whose length is less than r. An awful lot of them don't satisfy a.y=0. Can you find just one?
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