$T_1$-property of cofinite topology.

Question

Define $\tau$ on $\mathbb N$ by $\tau:=\{A\subset \mathbb N:\mathbb N-A $ is finite$\}(=\tau_f)$.Now I want to show that $\tau_f$ gives rise to a $T_1$ space on $\mathbb N$ although it is not $T_2$.Now take any two points $x,y$ in $\mathbb N$ which are distinct.Now assume $x<y$,consider $U_x=\{x,x+1,....\}-\{y\}$ and $U_y=\{y,y+1,....\}$.Then $U_x$ and $U_y$ are such that first contains $x$ and not $y$ and the second one vice versa.Since,$x,y$ are arbitrary,so the space is $T_1$,but it is not $T_2$ because it will give rise to a contradiction that $\mathbb N$ is finite.Is the proof correct?

Answer 1

The proof could be better, abstract away from the order altogether: take two distinct $x,y \in \Bbb N$. Then $\Bbb N - \{x\}$ is an open set (its complement is $\{x\}$ which is finite), contains $y$ and does not contain $x$. Similarly $\Bbb N - \{y\}$ is open and contains $x$ but not $y$.

The space is "anti-Hausdorff": any two non-empty open sets intersect: If $U,V$ are such sets, $F=\Bbb N - U$ is finite, $G=\Bbb N-V$ is finite and if $x \notin F \cup G$ (a finite set, so plenty left in $\Bbb N$) then $x \in U \cap V$ and so $U \cap V \neq \emptyset$. Make it concrete by saying e.g. $1$ and $2$ do not have disjoint open neighbourhoods, which is enough to refute Hausdorffness.

Answer 2

Looks good. Though maybe be a bit more explicit on why $T2$ gives a contradiction.

Also, as an exercise, try proving the same for any infinite set with the cofinite topology. On $\mathbb{R}$ or $\mathbb{C}$ this topology coincides with the Zariski topology.