Zariski Open Sets are Dense? Is it true that any nonempty open set is dense in the Zariski topology on $\mathbb{A}^n$? I'm pretty sure it is, but I can't think of a proof! Could someone possibly point me in the right direction? Many thanks!
Note: I am not asking about the Euclidean topology at all!
 A: As an exercise, let's reduce everything to statements about polynomials. Every open set contains a basic open set $U$, which is the complement of the zero set of some nonzero polynomial $f$, so it suffices to show that these are Zariski dense. The Zariski closure of a set is the intersection of the zero sets of all polynomials vanishing on it. This is equal to $\mathbb{A}^n$ if and only if any polynomial vanishing on $U$ vanishes on $\mathbb{A}^n$. Thus the claim is equivalent to the following statement about polynomials:

Suppose a polynomial $g$ has the property that if $f(x) \neq 0$, then $g(x) = 0$. Then $g(x) = 0$ for all $x$. 

But the condition is equivalent to the claim that $f(x) g(x) = 0$ for all $x$. Can you finish the problem from here? (Note that you need to assume $k$ infinite.) 
A: We have the following proposition:
Proposition: Let $X$ be a non-empty topological space. The following are equivalent:


*

*If we can write $X$ in the form $X=F\cup G$, where $F$ and $G$ are closed, then $X=F$ or $X=G$.

*If $U,V$ are two open sets of $X$ and $U\cap V=\emptyset$, then either $U$ or $V$ is empty.

*Any non-empty open set of $X$ is dense.


When one (hence any) of these conditions is satisfied, we say $X$ is irreducible.
Now, all that remains to show is that $\mathbb{A}^n$ is irreducible, which follows from the fact that its affine algebra is an integral domain.
A: A non-empty  topological space $X$ is said to be  irreducible if every non-empty open subset is dense.
Now there is a nice criterion for an affine scheme $X=Spec(A)$ to be irreducible (in the Zariski topology, of course): 
$$X \;\text {is irreducible} \iff A_{red} =A/Nil(A) \;\text {is a domain}  $$ 
So certainly if $k$ is a field, affine space $\mathbb A^n_k=\text {Spec}(k[T_1,...,T_n])$ is irreducible: every non-empty open subset $U \subset \mathbb A^n_k$ is  dense.
Edit: An elementary point of view
Since $\mathbb A^n(k)=k^n$ has a basis  of its topology given by the  $D(f)=\lbrace P\in k^n\mid f(P)\neq 0\rbrace \subset  k^n \; (f\in k[T_1,...,t_n])$, it is enough for proving irreducibility  of $\mathbb A^n(k) $  to see   that $D(f)\cap D(g)\neq \emptyset$  as soon as $D(f), D(g)\neq \emptyset$.
Equivalently  to see that $V(f)\cup V(g)=V(fg)\neq \mathbb A^n(k)$ (where  $V(f)=\mathbb A^n(k)\setminus D(f)$ etc.).
But this is indeed true as soon as the field $k$ is infinite: no non-zero polynomial can vanish at all points of  $k^n$.
Note carefully however that if  $k$ is finite, $k^n$ is discrete and thus $\mathbb A^n(k)$ is not irreducible.    
[I have sneakily turned from $\mathbb A^n_k$ to $ \mathbb A^n(k)    $ in the elementary treatment. In the scheme-theoretic point of view  $ \mathbb A^n_k    $ is irreducible even for finite $k$, as seen above  the Edit]
