The complement of a point in $\Bbb{A}^n$ is compact in the Zariski topology.

Question

The book "Invitation to Algebraic Geometry" says the following:

The complement of a point in $\Bbb{A}^n$ is compact in the Zariski topology.

Why is this this the case? This is thing that is asked in the commutative algebra section so I hope that Hilbert basis theorem might be used. Do we have to take a closure of the complement space?

Can we extend this to a question that any open cover of the complement of a variety will have a finite subcover?

Answer 1

This is going to sound crazy, but:

Every subset of $\mathbb{A}^n$ is compact.

The reason is almost purely set-theoretic, and here it is.

People call a topological space $X$ a noetherian space if it satisfies the ascending chain condition on open sets (equivalently, the DCC on closed sets). Notice $\mathbb{A}^n$ is noetherian, in fact any variety (irreducible or not) is noetherian --- this is basically because the coordinate ring is a finitely-generated algebra over a field, which is a noetherian ring, and closed sets in $\mathbb{A}^n$ correspond to ideals of these rings.

Here is the important exercise.

Great fact: $X$ is noetherian if and only if every subset of $X$ is compact.

Since $\mathbb{A}^n$ is noetherian, this answers your question!