Zariski topology questions from Atiyah and Macdonald's Introduction to Commutative Algebra Exercise 17 in Chapter 1 in Atiyah and Macdonald's Introduction to Commutative Algebra introduces the Zariski topology. There are 7 subquestions, of which 4 I've solved on my own, but the last couple are throwing me off. 

(iv) $X_f=X_g \Leftrightarrow r((f))=r((g))$, where $X_f$ signifies
  the basic open sets of the Zariski topology, otherwise defined as the
  complement $V(f)^c$, and "r" is for radical.

There's a ton of such mini questions on the Zariski topology and I've probably gotten a little tired of them because for some reason I can't get past this one!
Also, I proved that X is quasi-compact (=compact, really... they want to reserve compact for Hausdorff). But then the exercise reads:

vi) Specifically, each $X_f$ is quasi-compact.

I can't see why, I can't seem to extend my proof for $X$ quasi-compact to force each $X_f$ to be quasi-compact too.
Lastly,

vii) An open subset of $X$ is quasi-compact $\Leftrightarrow$ it is a
  finite union of sets $X_f$

Since the finite union of compact spaces is compact, the $<=$ direction is trivial. But the $=>$ direction seems a lot like part vi), I just don't see the connection between $X$ and $X_f$.
Any help is much appreciated. I'm trying to bone up my algebra knowledge before next semester.
 A: Hopefully this helps.  The moral is basically unwind the definitions and use a few facts about radicals over and over. 
$(iv)$: If $X_f = X_g$, then $V(f) = V(g)$ and so 
$$ r((f)) = \cap_{P\in V(f)} P =  \cap_{P\in V(g)} P = r((g)),$$
since the radical of an ideal is equal to the intersection of all the prime ideals that contain it.
On the other hand, if $r((f)) = r((g))$ we use the same property.  If $P \in V(f)$, then $(g) \subset r((g)) = r((f)) \subset P$ and so $P \in V(g)$. The other containment is the same.    
$(vi)$: Start with an arbitrary open cover of $X_f \subseteq \cup_{i\in I} X_{e_i}$. $X_f$ is open so we can assume this inequality is equality. Then $V(f) = V((e_i)_{i\in I})$.  This means, by definition, 
$$ f \in \cap_{P \in V((e_i)_{i\in I})}P = r((e_i)_{i\in I}). $$
Hence $f^n$ can be written as a finite sum $\sum_{j\in J} a_j e_j$  for some $n \in \mathbb{N}$ and so for every prime $P \in V((e_j)_{j \in J})$, we have $(f)\subseteq  P$. This shows $V((e_j)_{j\in J}) \subseteq V((f))$.  Lastly, take complements: 
$$X_f \subseteq X - V((e_j)_{j\in J}) = \cup_{j\in J} X - V((e_j)) = \cup_{j\in J} X_{e_j}. $$
$J$ is finite, so $X_f$ is compact.
$(vii)$: $(\Leftarrow)$ immediate from $(vi)$. $(\Rightarrow)$ If $K$ is any open compact subset of $X$, write it as a union of basic open sets, then take a finite subcover.
