The spectrum of a commutative ring with unity and its "topology" Let $\operatorname{Spec}(R)$ be the set of prime ideals in the commutative ring with unity $R$, and let $\mathfrak a$ be some ideal. Show that we get a topological space if we define the closed sets as the prime ideals in $R$ that don't include $\mathfrak a$.

What I tried
There are two things that have to be shown:


*

*An (uncountable) set of prime ideals $\{\mathfrak p_\alpha : \alpha \in I  \}$ such that 
$\mathfrak p_\alpha  \nsupseteq \mathfrak a$, satisfies $\bigcup_{\alpha \in I} \mathfrak p_\alpha \nsupseteq \mathfrak a$

*A finite set of prime ideals $\{\mathfrak p_1, \dots ,\mathfrak p_n \}$ such that 
$\mathfrak p_j  \nsupseteq \mathfrak a$, satisfies $\bigcap_{\alpha \in I} \mathfrak p_\alpha \nsupseteq \mathfrak a$


The last statement is clear, but I don't understand why the first one could be true. What if $\mathfrak a$ is "spread" over several sets $\{ \mathfrak p_\beta : \beta \in J \}$. In that case it wouldn't hold right?

I need a hint to go on. Thank you.
 A: The proposed "closed" subsets $D(\mathfrak a)$ consisting of the prime ideals not containing an ideal $\mathfrak a$ do not in general form a topology, because an arbitrary intersection of such subsets is no longer of the required form.    
Example:
Take $R=\mathbb Z$ and consider the "closed" subsets $D_p=D(p\mathbb Z)\subset \text {Spec}(\mathbb Z)$ for $p$ prime.
Their intersection is the singleton set $\{(0)\}\subset \text {Spec}(\mathbb Z)$.
 However that singleton set   $\{(0)\}$ cannot be of the form $D(\mathfrak a)$ for some ideal $\mathfrak a \subset \mathbb Z$. Why is that?
Because (remembering that ideals of $\mathbb Z$ are of the form $(n)$ with $n\in \mathbb N$):
a) If $\mathfrak a=(0)$, then $D(\mathfrak a)=\emptyset$
b) If $\mathfrak a=(1)$, then $D(\mathfrak a)=\text {Spec}(\mathbb Z)$
c) If $\mathfrak a=(n)$ with $n\gt 1$, then $D(\mathfrak a)$ contains all ideals generated by primes not dividing $n$ (and is thus infinite).
...and thus we'll never have $D(\mathfrak a)=\{(0)\}$.
