TITLE: A question about the Zariski topology
QUESTION [5 upvotes]: Let $(\mathfrak a_i)$ be an infinite family of ideals in commutative ring $R$. Is $\bigcap\limits_{i=1}^\infty \mathfrak a_i$ not defined?

I am trying to understand Zariski topology. Here, $V(\bigcap_i \mathfrak a_i)= \bigcup\limits_{i} V(\mathfrak a_i)$. 
If $\bigcap\limits_{i=1}^\infty \mathfrak a_i$ is defined, which I think it should be, then we would have an infinite union of sets of the form $V(X)$, all of which are closed sets. This is a contradiction as the infinite union of closed sets need not be closed.

REPLY [3 votes]: $V(\cap_i \mathfrak{a}_i)$ is the closure of $\cup_i V(\mathfrak{a}_i)$. In general, $\cup_i V(\mathfrak{a_i})$ might be not closed. For example, take $\mathbb{Z}$ so that $\mathrm{Spec}(\mathbb{Z})$ carries the cofinite topology except for the generic point $\eta$ which is contained in every non-empty open subet. We consider the ideals $(p)$ for prime numbers $p$. Then $\cup_p V(p) = \mathrm{Spec}(\mathbb{Z}) \setminus \{\eta\}$ is not closed (but it is dense in $\mathrm{Spec}(\mathbb{Z})$).