The complement of a zero-set of a polynomial is dense Let $P:\mathbb{R}^n\rightarrow \mathbb{R}$ be a non-zero $n$-dimentional polynomial. 
Let $C=\left \{ x\in \mathbb{R}^n: P(x) \neq 0 \right \} $.
How can I show that $C$ is dense?
I don't even know where to start.
 A: Let $a\in \mathbb R^n, r>0.$ We need to show $P(x)\ne 0$ for some $x\in B(a,r).$ Assume to the contrary that $P\equiv 0$ in $B(a,r).$ Then for every unit vector $u,$ the function of $t\in \mathbb R$ given by $q_u(t) = P(a+ tu)$ is $0$ for $t\in (-r,r).$ But $q_u(t)$ is a one-variable polynomial, and a one-variable polynomial that vanishes in a full interval vanishes identically. Thus each $q_u$ is identically $0.$ That is the same as saying $P\equiv 0$ on each line through $a.$ That implies $P\equiv 0$ on $\mathbb R^n,$ contradiction
A: Assume for the moment $n=2$.
Suppose the contrary. That is, assume that there is a disk $D(\mathbf x,r)$ where $P$ vanishes. Say that $\mathbf x=(a_0,b_0)$. Then the polynomial $P^2_{b_0}(x)=P(x,b_0)\in\Bbb R[x]$ has infinitely many roots, and then $P^2_{b_0}=0$. (The $2$ here is not a power, but just a superindex, and it points to what indeterminate is substituted by the subindex $b_0$).
That means that $b_0$ is a root of the polynomial $P^1_a(y)=P(a,y)\in\Bbb R[y]$ for every real $a$. That is, $P(a,y)=(y-b_0)Q^1_a(y)$ for every $a\in\Bbb R$, or
$$P(x,y)=(y-b_0)Q(x,y)$$
Since the degree of $Q$ is lesser than of $P$ we can use induction on the degree.
After this, we can use induction now on the number $n$ of indeterminates.
