Zariski Topology is Topology

Theorem
Let $k$ be a field.

Let $n \in \N_{>0}$.

Let $\tau$ be the Zariski topology on $k^n$.

Then $\tau$ is indeed a topology on $k^n$.

Proof
Let $A := k \sqbrk {x_1, \ldots, x_n}$.

Recall that by :
 * $\forall U \in \tau \; \exists T_U \subseteq A : U = X \setminus \map V {T_U}$

where $\map V {T_U}$ denotes the zero locus of $T_U$.

We shall show O1-O3 of.

O1
For each $\UU \subseteq \tau$:

02
For all $U,V \in \tau$:

O3
Since $\map V 1 = \O$:
 * $k^n = k^n \setminus \map V 1 \in \tau$