# Show that affine varieties are quasi-compact.

A topological space $X$ is called quasi-compact if whenever ${(U_i)}_{i∈S}$ are a family of open subsets such that $∪_{i∈S}(U_i) = X$ then there are a finite number of $(U_i)$’s which actually cover X.

I'm thinking first we show that if $(U_{fi})$, $i ∈ S$ (where ${f_i}$, $i ∈ S$ a collection of elements of $R[X]$) is a family of principal open subsets which cover an affine variety $X$, then there is a finite number which cover $X$.
Then we use the fact that principal open subsets are a basis for the Zariski topology and above statement to show that a finite number of the $(U_j)$ are sufficient to cover $X$.

• What is the definition of affine variety you are using? – user363120 Mar 7 '17 at 20:27
• Is an affine variety the set of zeros in $\Bbb A^n$ of a polynomial, or is it a reduced scheme of finite type? – Alex Mathers Mar 7 '17 at 20:45
• Its the set of zeros – user286826 Mar 7 '17 at 20:49
• You can see also here! – Armando j18eos Mar 8 '17 at 9:17

Every descending chain of Zariski closed subsets in $X$, $$X_1 \supseteq X_2,...$$

corresponds by inclusion reversal to $$I(X_1) \subseteq I(X_2)...$$

However, we know that Polynomial rings are Noetherian, so the ascending chain of ideals stabilizes.

Hence, every descending chain of closed sets does as well. By taking complements, we see that every ascending chain of open subsets stabilizes.

Let $\{B_i\}$ be some open covering of $X$. Consider the open sets that can be written as the finite union of elements in our open cover:$$\mathcal{A}:=\{ U \mid U=\bigcup_{i=1}^{k}B_i\}.$$

By the axiom of dependent choice, we know that if $\mathcal{A}$ does not contain $X$, then there exists some infinite ascending chain of open sets, which is a contradiction.

Affine schemes are Noetherian as topological spaces, meaning every descending chain of closed subspaces $Z_1\supseteq Z_2\supseteq\cdots$ stabilizes. This is the same as saying that each increasing chain of open subspaces $U_1\subseteq U_2\subseteq\cdots$ stabilizes.

Now try to show that if $X$ is not quasicompact, you can get an infinite strictly increasing chain of open subsets, giving a contradiction.