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

• werdna91
If you assume x is in S, then you can find a y that is in B(x,r) and also not in S.In summary, the conversation discusses a problem that asks to prove that the interior of a set S, which is defined as the set of all orthogonal vectors to a non-zero vector "a" in R^n, is empty. The solution involves showing that S is its own boundary, and using the fact that a vector x in S must have a dot product of 0 with any vector in the complement of S. The proof considers an example of a vector y in the complement of S that is also in the open ball B(r, x), and shows that this leads to a contradiction, proving that the interior of
werdna91

## Homework Statement

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

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.

Last edited by a moderator:
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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Is my statement true:

A set which interior is empty is a surface.

Is that true ?

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.

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

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?

werdna91 said:
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.

## 1. What does it mean for a set of vectors to be orthogonal?

Two vectors are orthogonal if their dot product is equal to zero. In other words, they are perpendicular to each other.

## 2. How can I prove that the interior of the set of all orthogonal vectors to a is empty?

In order to prove that the interior of a set is empty, we need to show that there are no points within the set that are also contained in a larger open set. In this case, we need to show that there are no orthogonal vectors to a that also lie within a larger open set.

## 3. Can you provide an example of an orthogonal vector to a?

Yes, an example of an orthogonal vector to a would be any vector that is perpendicular to a. For example, if a is a vector in the xy-plane with components (3, 0), then any vector in the z-direction (0, 0, z) would be orthogonal to a.

## 4. Why is it important to prove that the interior of the set of all orthogonal vectors to a is empty?

This proof is important because it helps us understand the properties of orthogonal vectors and their relationship to other vectors. It also allows us to make conclusions about the geometry of the vector space.

## 5. Are all sets of orthogonal vectors empty?

No, not all sets of orthogonal vectors are empty. It is possible to have a set of orthogonal vectors that contains a single vector or multiple vectors. However, the interior of the set of all orthogonal vectors to a will always be empty.

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