TITLE: Prove or disprove the continuity of the Identity function on the topological space?
QUESTION [5 upvotes]: Let $\tau_1$be the usual topology on $\mathbb R$ . Define another topology $\tau_2$ on $\mathbb R$ by
$$\tau_2 = \{U \subseteq \mathbb R \ \ | \ \ U^c \ \ is \ \ either \ \ finite \ \ or \ \ empty \ \ or \ \mathbb R\ \ \}$$
If $I : (\mathbb R , \tau_1) \rightarrow (\mathbb R , \tau_2)$ is the identity map , then 

$I$ is continous but not $I^{-1}$
$I^{-1}$ is continous but not $I$.
both $I$ and $I^{-1}$ are continous .
neither $I$ nor $I^{-1}$ are continous .

My attempt is :
since $(0,\infty)$ is open in $\tau_1$ but not in $\tau_2$ . so $\tau_2$ is not finer than $\tau_1$ . so $I^{-1}$ is not continous because we have a result $I$ is continous iff $\tau_1$ is finer than $\tau_2$ , similarly this result hold for $I^{-1}$
I think $I$ is continous because take any closed set in $\tau_2$ which is a finite set is closed in $\tau_1$ because R is Hausdorff.
Iwould be thankfull who someone give me your valuable time  to check my solution.

REPLY [5 votes]: Since $I$ is an identity map, it's really simple.
Just ask 2 questions:
(1) Is every set which is open in $\tau_2$ also open in $\tau_1$?
(2) Is every set which is open in $\tau_1$ also open in $\tau_2$?
The first is true iff $I$ is continuous and the second is true iff $I^{-1}$ is continuous.
Your attempt seems to be correct as $(0, \infty)$ is open in $\tau_1$, but $(-\infty,0]$ is not finite and any set open in $\tau_2$ is the complement of a finite set, but finite sets are closed in $\tau_1$, so the complement is open.
So yes, $\tau_1$ is finer than $\tau_2$. Also, you need only that $\mathbb R$ is a $T_1$ space to know finite sets are closed, Hausdorff ($T_2$) is stronger than that.