Closed sets in Spec(k[X,Y]) On page 74 of Mumford's red book (attached) it is stated that a proper closed set in Spec(k[X,Y]) is composed of finitely many irreducible curves and finitely many closed points. Why does such a union need to be finite? Also, he states that given a closed subset of the set of closed points in the classical Zariski topology, there is a unique set of non-closed points to add to get a closed set in the new topology. Is a closed subset of the set of closed points in the classical topology not closed in the modern? Thanks!
 A: To see why it needs to be a finite union:
Let $A$ be a closed set so that $A=\bigcup V(I_i)$ for ideals $I_i$. Then $A=V(\bigcap I_i)$ by properties of the Zariski topology. Since $k[x,y]$ is Noetherian, there are only finitely many minimal primes (call them $P_j$) over $\bigcap I_i$ so that this reduces to the case of $A=V(P_1)\cup \cdots \cup V(P_n)$.
For the second question, it would probably be easier to think of what Mumford is saying like this:
Clearly the ring is invariant in the correspondence between the classical affine space and Zariski topology and the spectrum and the Zariski topology on the spectrum. So if we take a closed subset of affine space, the closed points of this set correspond to a set of maximal ideals in our ring. If we now consider this set of maximal ideals in our ring, but corresponding to points in the spectrum, the closure of this the points of the spectrum includes new points (non-closed, generic points).
A: (In this answer I will primarily use $k=\mathbb{C}$, since I think $\mathbb{C}$ gives us a better "picture", being more familiar)
You seem to be confusing two topologies on $\mathbb{C}^2$. By "classic topology", I assume you mean the standard topology induced by the standard metric. This is different from the Zariski topology (which I assume you meant by "modern topology"). As you point out $\mathbb{C}^2$ with Zariski topology is very different from $\mathbb{C}^2$ with the standard topology you might be used to. To get a sense of Zariski topology, you should think that open sets are VERY big in Zariski topology, whatever that means. 
For example, let's take $\operatorname{Spec}\mathbb{C}[x]$, which you can think of as $\mathbb{C}$ (Here, it is important to remember that the closed points of $\operatorname{Spec}k[x]$ are identified with $\mathbb{C}$ since $\mathbb{C}$ is algebraically closed). Then, in standard topology, closed sets are finite unions of closed intervals, or the entire $\mathbb{C}$. However, a closed set in Zariski topology is simply a finite set of points. Eoin's answer explains why and another easy way to see this is by noticing that a polynomial $f\in \mathbb{C}[x]$ has only finitely many zeros. Since $\mathbb{C}[x]$ is a PID, any maximal ideal is generated by some $f$ and thus, closed sets are zeros of polynomials. This means that Zariski open sets of $\operatorname{Spec}\mathbb{C}[x]$ are precisely those whose complement is a finite set of points.
The case of $\operatorname{Spec} k[x,y]$ is similar to the case above. The Zariski closed sets are going to be finite unions of loci where the ideal "vanishes" i.e. curves and closed points. 
NOTE The statement is true even when $k$ is not algebraically closed. However, if $k$ is not algebraically closed, $\operatorname{Spec}k[x,y]$ does not "look like" $k^2$. More precisely, the closed points of $\operatorname{Spec}k[x,y]$ do not correspond to points of $k^2$. You can try seeing what the closed points of $\operatorname{Spec}\mathbb{R}[x]$ are and it will give you a good idea. I hope this helps!
A: Consider the Zariski topology on $\mathbb{A}^2_k$ for $k$ algebraically closed. Then irreducible closed subsets are  $\mathbb{A}^2_k, V((f))$ for $f$ irreducible and $V((x-a,y-b))$. Thus, the closed subsets are the whole space, finite collection of points and irreducible curves. 
An arbitrary closed subset in $\mathbb{A}^2_k$ just looks like  a bunch of points in $\mathbb{A}^2_k$. 
This is the old topology. 
By considering $\mathbb{A}^2_k \subset spec(k[x,y])$ we realise the bunch of points in $\mathbb{A}^2_k$ is not a closed set in $spec(k[x,y])$. What we need to do to get closed subsets is just add the non-closed points. This exactly corresponds to addings points like $(f)$ or $(0)$ (thinking of them as elements in $spec(k[x,y]) )$.  

