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 $A := k \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomials.

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

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

Each of the open set axioms is examined in turn:

For each $\UU \subseteq \tau$:

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

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

All the open set axioms are fulfilled, and the result follows.