I am trying to complete this Topological problem and I have completed it, I just would like some opinions on how to make my work better and if someone could check it for me as well. I would greatly appreciate it!
(A): Declare a set $U \subseteq \Re $ to be "open" if its complement $\Re $\ $U$ is the zero-set of a (real-valued) polynomial $p$ $\in$ $P$ ($\Re$):
$U$ is open $\Leftrightarrow$ $\exists$$p$ $\in$ $P$ ($\Re$) such that $\Re$\ $U$ = $p$ $^-$$^1$$(0)$.
Show that the collection of all "open" sets is a topology on $\Re$.
MY SOLUTION TO PART A:
Let $U$$_1$,$U$$_2$$...$$U$$_n$ be an arbitrary collection of open sets. Then $\Re$ \ $U$$_1$,$\Re$ \ $U$$_2$, $...$ $\Re$ \ $U$$_n$ would be the zero set of a polynomial $P$$_1$,$P$$_2$,$...$$P$$_n$ respectively. Now let $U$ = $\cup^{\infty}_{i=1}U_\Re$. As $\Re$ \ $U$ $\subseteq$ $\Re$ \ $U$$_1$, we get that $\Re$ \ $U$ is the zero set of a polynomial which divides $P$. Now, suppose that $V$$_1$,$V$$_2$,$...$,$V$$_n$ are open sets and $V$ = $V$$_1$ $\cap$ $V$$_2$ $\cap$ $...$ $\cap$ $V$$_n$. Then we get that $\Re$ \ $V$$_1$,$\Re$ \ $V$$_2$, $...$ $\Re$ \ $V$$_n$ are the zeros of a polynomial $P$$_1$,$P$$_2$,$...$$P$$_n$. We also get that $\Re$ \ $V$ are the roots of a polynomial $P$$_1$,$P$$_2$,$...$$P$$_n$. Thus, $V$ is open. Since open sets are closed under arbitrary unions and finite intersections, they form a topology.
(B): Show that the topology in (a) is equal to the cofinite topology.
MY SOLUTION TO PART B:
Let $U$ be an open set
$\Leftrightarrow$ $\Re$ \ $V$ is the zero set of a polynomial since every polynomial only has finite roots.
$\Leftrightarrow$ $\Re$ \ $V$ is a finite set
$\Leftrightarrow$ $V$ is an open set in the cofinite topology. Thus, the topology in (a) is equal to the cofinite topology.
(C): Prove weather the topology is Hausdorff or not.
MY SOLUTION TO PART C:
Let V be an open set $\Rightarrow$ $\Re$\V is finite, so if $\Re$=$U$$\cap$$V$ and if $U,V$ are distinct, then $V$ is finite. But, all open sets in the cofinite topology contain infinitely many elements. Thus, the topology is not Hausdorff