$V_{\mathbb{A}^n}(\sum_{\imath}\mathfrak{a}_{\imath}) = \bigcap_{\imath} V_{\mathbb{A}^n}(\mathfrak{a}_\imath)$ Let
$$
V_{\mathbb{A}^n}(\mathfrak{a}) := \{x \in \mathbb{A}^n \; \vert \; f(x) = 0 \; \text{for all} \; f \in \mathfrak{a}\}, 
$$
for some affine space $\mathbb{A}^n$, a polynomial $f$ and an ideal $\mathfrak{a}$. I want to proof that
$$
V_{\mathbb{A}^n}(\sum_{\imath}\mathfrak{a}_{\imath}) = \bigcap_{\imath} V_{\mathbb{A}^n}(\mathfrak{a}_\imath),
$$
and that
$$
V_{\mathbb{A}^n}(\bigcap_{i=1}^{n} \mathfrak{a}_i) = \bigcup_{i=1}^{n}V(\mathfrak{a}_i).
$$
How can I prove above equations?
And why is the finite case in the second one so important?
Thoughts on the first case: The sum of two ideals is again an ideal as an abelian group can be subdivided into subgroups or joint under the group operation to a larger abelian subgroup. Multiplication with an element of the ring preserves still the group structure, thus yields again an element of the abelian group.
 A: This is Hartshorne Lemma II.2.1(b), but I appreciate that Hartshorne's proof is a little sparse in terms of details.
As a set $V_{\mathbb{A}^n}(\mathfrak{a})$ is the set of all prime ideals containing $\mathfrak{a}$, so $V_{\mathbb{A}^n}(\sum \mathfrak{a}_i)$ is the set of all prime ideals containing $\sum \mathfrak{a}_i$. Recall that $\sum \mathfrak{a}_i$ is the smallest ideal containing every $\mathfrak{a}_i$, so an ideal $\mathfrak{p}$ contains $\sum\mathfrak{a}_i$ if and only if $\mathfrak{p}$ contains each $\mathfrak{a}_i$. Thus the set $V_{\mathbb{A}^n}(\sum_i \mathfrak{a}_i)$ of prime ideals containing $\sum_i \mathfrak{a}_i$ is the same as the set $\bigcap_i V_{\mathbb{A}^n}(\mathfrak{a}_i)$ of prime ideals contatining each $\mathfrak{a}_i$.
The second one is similar and I encourage you to try and prove it - once you have the $n = 2$ case you're done by induction, so try to understand why $V(\mathfrak{ab}) = V(\mathfrak{a}) \cup V(\mathfrak{b})$; if you have a prime ideal containing $\mathfrak{ab}$ can you show that it contains either $\mathfrak{a}$ or $\mathfrak{b}$ (think about why we used a prime ideal here).
The finite case is important because of the definition of topological spaces; the set of opens is closed under arbitrary union but closed under only finite intersections. This statement is saying that the $V_{\mathbb{A}^n}(\mathfrak{a})$ form a base for a topology (the Zariski topology).
edit: as per the reccomendation of LetGBeTheGraph in the comments, let's see the example of why the union of every prime ideal of $\mathbb{Z}$ is not closed in this Zariski topology.
Recall that $n \in (k)$ iff $n$ is a multiple of $k$. Thus for example the intersection $(2) \cap (3) = (6)$ because if $n \in \mathbb{Z}$ is a multiple of two and three then it's a multiple of six. Thus any $a \in \bigcap_{p\text{ prime}} (p)$ must simultaneously be a multiple of every prime number, no such integer satisfies this, so there is no prime ideal containing $\bigcap_{p\text{ prime}} (p)$, in our terminology; $V_{\mathbb{Z}}(\bigcap_{p\text{ prime}} (p)) = \emptyset$.
We compare this to $\bigcup_{p\text{ prime}} V_{\mathbb{Z}}((p))$. Given any prime $p$ we have that $(p)$ is a prime ideal containing $(p)$, i.e. $(p) \in V_{\mathbb{Z}}((p))$, so at very least we have that $(p) \in \bigcup_{p\text{ prime}} V_{\mathbb{Z}}((p))$. Thus we have that $\bigcup_{p\text{ prime}} V_{\mathbb{Z}}((p)) \neq V_{\mathbb{Z}}(\bigcap_{p\text{ prime}} (p))$.
