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I made a picture of the problem here: (statement of the problem, rendered as an image)

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Let $a$ be a non-zero vector in $\mathbb{R}^n$. Let S be the set of all orthogonal vectors to $a$ in $\mathbb{R}^n$. I.e., for all $x \in \mathbb{R}^n$, $a\cdot x = 0$

Prove that the interior of S is empty.

How can I show that for every point in S, all "close" points are either in the complement of S or in S itself?

This is what I attempted:

Let $u\in B(r,x) = \{ v \in \mathbb{R}^n : |v - x| < r \} $

So $|u - x| < r$

Then, $|a||u - x| < |a|r$.

By Cauchy-Schwarz, $|a\cdot(u-x)| \leq |a||u - x|$.

Then, $|a\cdot u - a\cdot x| < |a|r$.

If $u\in S$, then either $u\in S^{\text{int}}$ or $u\in \delta S$. ($\delta$ denotes boundary).

If $u\in S^{\text{int}}$, then $B(r,x) \subset S$, and $a\cdot u = 0$. But then the inequality becomes $|a|r > 0$ which implies $B(r,x) \subset S \forall r > 0$, but this is impossible since it would also imply that $S = \mathbb{R}^n$ and $S^c$ is empty, which is false. Therefore, if $u\in S$, then $u\in\delta S$.

Hence, $\forall u\in B(r, x)$ such that $u\in S$, $u\in\delta S$. Thus $S^{\text{int}}$ is empty.

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2  
Is this homework? –  Aryabhata Sep 28 '10 at 20:59
    
I added my attempt to the solution. Is it the right direction? Thanks. –  user2193 Sep 28 '10 at 21:51
    
One approach would be to show projection is an open map. In particular the projection of an open ball is another open ball. –  Ryan Budney Sep 28 '10 at 22:03

4 Answers 4

up vote 1 down vote accepted

It is a tautology to say "every point in S, all 'close' points are either in the complement of S or in S itself". Also I am not sure how you use $a|r|>0$ to conclude that $B(r,x) \subset S \, \forall r>0$.

Anyway you don't need to think of all "close" points. You just need to prove that arbitrarily close to any $x \in S$ there is at least one point which is not in $S$.

Hint: think of points of the form $x+\lambda a$.

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Thanks, I think this was the best way to prove it. I ended up taking $\lambda < \frac{r}{|a|}$. I showed that $y = x + \lambda a$ is in the complement of $S$, and then I showed that $y$ was in $B(r, x)$. This implies that $x$ is a boundary point, so for all $x\in S$, $x$ is a boundary point, and so the interior of S is empty. –  user2193 Sep 29 '10 at 15:01

Consider the function $\phi:x\in\mathbb R^n\mapsto x\cdot a\in\mathbb R$, which is clearly linear. If $x_0$ is a point in the interior of your set, compute the derivatives of $\phi$ at $x_0$. Now look at the Taylor development of $\phi$ at $x_0$...

A more geometric way: suppose that $x$ is orthogonal to $a$ (and that $a\neq0$) and consider the sequence $(x_n)_{n\geq1}$ with $x_n=x+\tfrac1na$ for each $n$. Show that $x_n\to x$ when $n\to\infty$ and then see why this is useful for you.

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I would do it this way. First of all, by choosing $\frac{a}{\|a\|}$ as the last vector of an orthonormal basis of $\mathbb{R}^n$, you may assume that $a = (0,\dots , 0,\alpha), \alpha \in \mathbb{R}, \alpha \neq 0$. Hence

$$ S =\left\{ (x_1 , \dots , x_n) \in \mathbb{R}^n\ \vert \ x_n = 0 \right\} \ . $$

Now, pick any $x = (x_1 , \dots , x_{n-1} , 0) \in S$ and let's show that, for any $r>0$, $B(r,x) $ is not included in $S$, so the interior of $S$ will be empty.

To this end it's enough to produce a single vector in $B(r,x)$ which is not in $S$, right? Ok, take a look at $v = (x_1, \dots , x_{n-1}, \frac{r}{2})$ and compute (draw a picture in $\mathbb{R}^3$ too to convince yourself).

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The condition that $x$ be orthogonal to $a$, i.e. that $x$ lies in $S$, is that $x \cdot a = 0$. Imagine perturbing $x$ by a small amount, say to $x'$. If $x$ were in the interior, than one would have $x' \cdot a = 0$ as well, provided that $x'$ is very close to $x$.

Think about whether this is possible for every $x'$. (Hint: $x'$ has to be close to $x$, i.e. $x - x'$ has to be small. But it can point in any direction!)

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