Show that $\mathbb{A}^n$ on the Zariski Topology is not Hausdorff, but it is $T_1$ There was an exercise I could not do. So the property is $T_1$ if for every pair of distinct points, $P, Q \in X$, there is an open subset $U$ containing $P$ but not $Q$ and another open subset $V$ containing $Q$ but not $P$.
So I was asked to show if the base field is infinite, then $\mathbb{A}^n$ is $T_1$ but not Hausdorff.
I can show that it's not Hausdorff (I believe this is right). So let $k$ be an infinite field, then we have $U$ and $V$ which are two nonempty subsets which are open, thus $U^c$ and $V^c$ finite so by Demorgan's law we have $(U \cap V)^c =U^c \cup V^c$ which is also finite, so this implies that $U\cap V$ is non-empty so it can't be Hausdorff.
But I can't seem to really show that it's $T_1$
Any ideas how to approach this?
Thank you.
 A: Since Seth has proved that $\mathbb A^n$ is $T_1$, I'll prove that $\mathbb A^n$ is not Hausdorff [Seth correctly pointed out that Jen's proof is incorrect for $n\gt1$] .  
By considering complements of open subsets Hausdorffness amounts to proving that the union of two closed strict subsets $X_1\subsetneq \mathbb A^n, X_2\subsetneq \mathbb A^n$ is not the whole affine space: $X_1 \cup X_2\subsetneq \mathbb A^n$.
Now, writing $X_i=V(I_i)$, the inequality $X_i \neq \mathbb A^n$ is equivalent to $f_i\in I_i$ for some non zero $f_i\in k[T_1,\dots,T_n]$.
We have $X_1\cup X_2=V(I_1\cdot I_2)\subset V(f_1\cdot f_2)$, so that it is enough to prove $V(f_1\cdot f_2)\neq \mathbb A^n$  to obtain the desired conclusion $X_1 \cup X_2\neq \mathbb A^n$.
But $V(f_1\cdot f_2)\neq \mathbb A^n$ is clear: it results from the well-known fact that if $k$ is an infinite field a non-zero  polynomial     $g(T_1,\dots,T_n)\in k[T_1,\dots,T_n] $ (here $g=f_1\cdot f_2$) cannot vanish on all $(a_1,\dots,a_n)\in k^n$ .
A: $T_1$ is equivalent to single point sets being closed.  
Let $(a_1,\dots,a_n)\in\mathbb{A}^n$.  Consider the zero locus of the polynomials $x_i-a_i$ where $i$ ranges from $1$ to $n$.
For your proof of non-Hausdorffness it looks like you are assuming that $\mathbb{A}^n$ has the cofinite topology, which is not true in general.  $\mathbb{A}^1$ does have the cofinite topology, but $\mathbb{A}^2$ does not, since for example, the polynomial $x\in\mathbb{k}[x,y]$ has infinitely many zeros.  
