# Point Set Topology (open sets, polynomials, hausdorff spaces)

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 $$\Leftrightarrow$$ $$\existsp$$ $$\in$$ $$P$$ ($$\Re$$) 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 $$\Re$$=$$U\capV$$ 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

• Wouldn't it be easier to observe that the zeo-sets of polynomials are precisely the finite sets? – Paul Frost Apr 16 at 23:09
• To be more precise: You use this fact to prove (B). But then you can use it also in part (A). – Paul Frost Apr 17 at 15:21
• Yeah, that would be easier. I didn't think of that. However, are my solutions correct? – DataD96 Apr 17 at 20:57
• 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 $U_\alpha$. – Paul Frost Apr 17 at 21:55
• 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. – M. Winter Apr 23 at 19:37