Prove that the interior of the set of all orthogonal vectors to a is empty.

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  • #1
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Homework Statement



Here is a picture of the problem:
http://img84.imageshack.us/img84/1845/screenshot20100927at111.png [Broken]

If the link does not work, the problem basically asks:

Let "a" be a non-zero vector in R^n. Let S be the set of all orthogonal vectors to "a" in R^n. I.e., a•x = 0 (where • denotes dot product).

Prove that the interior of S is empty.

Homework Equations



B(r, a) = { x in R^n : |x - a| < r }

The Attempt at a Solution



I realize that I basically have to show that S is its own boundary. In other words, given a vector x in S, I must show that for every open ball B(r, s) (centred at s with radius r > 0), it contains points that are in S and points that are in the complement of S.

My trouble lies with somehow linking dot product with distance .

So if I say, let x be in S, and let y be in S^c (the complement of S).

How can I show that for all r > 0, y is in B(r, x)? All I know is that a•y ≠ 0. Clearly the norm of the vectors are irrelevant to their existence in the set S. But norm obviously has something to do with their existence in B(r, x).

Perhaps I'm over-thinking this. Clarification and suggestions for next steps would be great.

Thanks a lot.
 
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Answers and Replies

  • #2
Dick
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Yes, probably overthinking. If x.a=0, try to find an example of a vector y which is in B(x,r) for which y.a is not zero. Hint: a.a isn't zero.
 
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  • #3
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I haven't got an answer but I thought about the problem.
Is my statement true:

A set which interior is empty is a surface.

Is that true ?
 
  • #4
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Quinzio, I believe that is true. If S^int is empty, then S = the boundary of S.
Here is the proof I came up with, can somebody verify it?


Given S, we know that S^c = { y in R^n : a•y ≠ 0 } which is not empty.

Let x be in S and suppose that its interior is not empty.

Then there exists an r > 0 such that B(r, x) is in S.

Let u be in B(r, x), then u is an interior point of S.

We have that |u - x| < r for some r > 0.

So |a||u - x| < |a|r

By Cauchy-Schwarz, we have |a•(u - x)| ≤ |a||u - x|
Therefore, |a•(u - x)| < |a|r

Then |a•u - a•x| < |a|r.

We know that a•x = 0. Because u is an interior point of S, u is in S, so a•u = 0. But this yields 0 < |a|r, which holds for all r > 0. So no matter how big r is, B(r, x) is in S. This implies that S = R^n since all balls centered at any point in S is also in S. But this would also imply that S^c is empty, which is impossible. Therefore, a•u ≠ 0 which would contradict that u is in the interior of S, and therefore, B(r, x) is not in S for all r > 0. Hence, the interior of S is empty.
 
  • #5
Office_Shredder
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I don't think that what you said works as written. You started by assuming that B(x,r) is in S, and then through a series of steps proved that this implies that |a|r>0. This does not immediately imply the other direction: that if a|r|>0, B(x,r) is in S. Not all your steps are if and only if statements (for example, you used the fact that u is in B(x,r) to get a•u=0, but a•u=0 does not imply u is in B(x,r)) so you can't just turn the statement around like that
 
  • #6
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But by assuming that B(r, x) is in S, I thought I could conclude that a•u=0?

Do you think there is a better approach for me to prove this?
 
  • #7
Dick
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But by assuming that B(r, x) is in S, I thought I could conclude that a•u=0?

Do you think there is a better approach for me to prove this?

If you ASSUME B(x,r) is in S, then sure a*u=0. But so what? B(x,r) ISN'T in S. You want to prove there is a point in B(x,r) that ISN'T in S. Take another look at my post 2.
 

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