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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

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    $\begingroup$ Wouldn't it be easier to observe that the zeo-sets of polynomials are precisely the finite sets? $\endgroup$ – Paul Frost Apr 16 at 23:09
  • $\begingroup$ To be more precise: You use this fact to prove (B). But then you can use it also in part (A). $\endgroup$ – Paul Frost Apr 17 at 15:21
  • $\begingroup$ Yeah, that would be easier. I didn't think of that. However, are my solutions correct? $\endgroup$ – DataD96 Apr 17 at 20:57
  • $\begingroup$ Your proof is essentially correxct. The only point I would criticize is your proof that unions of open sets are open. You have that start with an arbitrary (not necessarily finite or denumerable) collection of open $U_\alpha$. $\endgroup$ – Paul Frost Apr 17 at 21:55
  • $\begingroup$ The construction here is similar enough to the Zariski topology that I wonder whether the question really restricts to polynomials in only a single variable. $\endgroup$ – M. Winter Apr 23 at 19:37

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