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Interior and boundary of set of orthogonal vectors

  • Thread starter trap101
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Let "a" be a non zero vector in R^n and define S = { x in R^n s.t. "a" · "x" = 0}. Determine S^int , bkundary of S, and closure of S. Prove your answer is correct


Attempt:

Ok im more sk having trouble proving that the respective points belong to its condition. Such as thr interior. I know will be all points in thr plane of X, but how can i show that all thise points are the interior if S? All i can think of is letting a point, call it "y", be in thr B(r,a) and showing that it is a interior point through the triangle inequality manipulation. But that doesn't use the fact of orthogonality in the set.
 

Answers and Replies

  • #2
HallsofIvy
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If a is a single vector in Rn, then the set of all vectors that are orthogonal to a is an n-1 dimensional "plane". You shouldn't have to use any "neighborhoods" or other technical details.

When in doubt, look at simple variations of the problem. Suppose n= 2 and a= <1, 0>. The the set of all vectors orthogonal to a is {<0, y>} for any y- the entire y- axis. What are the boundary, interior, and closure of the y- axis? Suppose n= 3 and a= <0, 0, 1>. The set of all vectors orthogonal to a is {<x, y, 0>} for any x and y- the xy-plane. What are the boundary, interior, and closure of the xy-plane?
 
  • #3
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ok, well using the examples you gave. The boundary, interior, and closure for n= 2: y- axis is the boundary, empty interior, and y axis is the points in the closure.

Now I tried extending it to n-1 imagining a plane and I got a similar result I believe:

Boundary would be all points in the n-1 plane, interior would be empty, and the closure would be all the boundary points
 

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