Prove that the interior of the set of all orthogonal vectors to "a" is empty I made a picture of the problem here:

If the link does not work, read this:
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.
 A: 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.
A: 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 $x+\lambda a$.
A: 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).
A: 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!)
