Trivial question with respect to the compactness of affine varieties under Zariski topology.

To show that Zariski topology on an affine algebraic variety is compact, I know that we can use the Noetherian property and construct a chain of ideals which stablize.

I don't know what is wrong with my proof though:

Let $V\subset \cup U_i$ be a variety that is covered by the union of open sets $U_i$. Then $A_n \backslash V \supset \cap (A_n \backslash U_i)$. Now since $A_n \backslash U_i$ are closed sets, their intersection is a closed set, which is an affine variety $\tilde{V}$. So $V\subset (A_n\backslash \tilde{V})$, which means $V$ is covered by one open set.

What is wrong with my argument? Thanks for your help!

• Nothing is wrong with your argument, but it doesn't prove what you want it to. You need to prove the original cover has a finite subcover, not just that $V$ is covered by a finite number of open sets (which is true of every subset of every topological space!) The open set $A_n \setminus \overline{V}$ is not an element of your original cover. – user64687 Feb 4 '15 at 11:29
• Oh, I see! Thank you! – KittyL Feb 4 '15 at 11:31

Why should $A_n \setminus \bar V$ arise under the $U_i$? What i say is: This is definitely a cover, but not a subcover of the given cover.