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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

  1. $I$ is continous but not $I^{-1}$

  2. $I^{-1}$ is continous but not $I$.

  3. both $I$ and $I^{-1}$ are continous .

  4. 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.

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    $\begingroup$ It looks fine. You can argue your first point directly, though: $(I^{-1})^{-1}( (0,\infty))$ is not open. $\endgroup$ – David Mitra Feb 28 '14 at 13:15
  • $\begingroup$ Is the result "I is continuous iff $ \tau_1 $ is finer than $ \tau_2 $" true for all topological spaces? I mean suppose $(X,\tau_1)$ and $(X,\tau_2)$ were any topological spaces then could this result be applied? $\endgroup$ – Error 404 May 23 '15 at 7:21
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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.

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