Let $X$ be $\mathbb{C}$ with the following topology: A subset $F$ of $X$ is closed $\iff$ $F$ is the set of zero(s) of a polynomial. Connected?

Question

Let $X$ be $\mathbb{C}$ with the following topology: A subset $F$ of $X$ is closed $\iff$ $F$ is the set of zero(s) of a polynomial.

Is $X$ Compact? Connected?

$X$ is indeed Compact. If $\{U_\alpha\}$ is an open cover of $X$, then the complement of each one is finite. Thus we can take one to cover all but a finitely many points of $X$, and then choose a finite number of elements of the cover to achieve a finite cover.

As for it being connected... A space $X$ is connected $\iff$ there is no continuous onto map $f: X \rightarrow \{0,1\}$ where $\{0,1\}$ has the discrete topology.

I think there does not exist such a continuous map... To have the inverse image of both $0$ and $1$ be open would mean that both of their preimages have finite complements, which is not possible. $\therefore X$ is connected.

Answer 1

Suppose that $X$ is the union of the non empty disjoint open subsets $U$ and $V$, the complementary of $U$ is $V$, which is also closed, thus $V$ is finite, similarly $U$ is closed since it is the complementary subset of $V$ so it is finite contradiction since $\mathbb{C}$ is not finite.

Answer 2

Here’s another proof.

Let $U_1$ and $U_2$ be any two (nonempty) open sets in this topology, with complements $F_1$ and $F_2$. Since

$$U_1 \cap U_2= \mathbb{C} \setminus (F_1 \cup F_2)$$

and because $F_1$ and $F_2$ are finite, this intersection cannot be empty. Therefore, no two open sets can ever be disjoint. This proves that the given topology is both connected and non-Hausdorff.