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$f(x)=x$.

$X$ is the set of all real numbers with finite complement topology (A set is open in this space iff it's complement is finite).

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  • $\begingroup$ Ok, I assume $X$ is just $\mathbb E ^1 = \mathbb R$ in the finite complement topology. $\endgroup$ – Forever Mozart Jun 25 '14 at 15:17
  • $\begingroup$ What is $\mathbb{E}^1$? Is it a fancy way to denote $\mathbb{R}$ with the standard topology (one-dimensional Euclidean space)? $\endgroup$ – Daniel Fischer Jun 25 '14 at 15:18
  • $\begingroup$ @TomCruise I have edited the question. $\endgroup$ – Mathew George Jun 25 '14 at 15:20
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    $\begingroup$ But the cofinite topology is (strictly) coarser than the standard topology, so $f^{-1}(\text{open})$ is open. $\endgroup$ – Daniel Fischer Jun 25 '14 at 15:38
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    $\begingroup$ Let $\tau_1,\tau_2$ be two topologies on the same set. Then $\tau_1$ is coarser than $\tau_2$ (and $\tau_2$ is finer than $\tau_1$) if and only if $\tau_1 \subset \tau_2$. If the topologies are not identical, then $\tau_1$ is strictly coarser than $\tau_2$, and $\tau_2$ strictly finer than $\tau_1$. $\endgroup$ – Daniel Fischer Jun 25 '14 at 15:55
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This follows from a general fact: If $\mathcal{O}$, $\mathcal{O}^\prime$ are two topologies on a set $X$ such that $\mathcal{O}$ is finer than $\mathcal{O}^\prime$ (i.e., every set in $\mathcal{O}^\prime$ is also in $\mathcal{O}$), then the identity function $\mathrm{id}_X : x \mapsto x$ is continuous as a function from $\langle X , \mathcal{O} \rangle$ to $\langle X , \mathcal{O}^\prime \rangle$. (In fact, the converse also holds: if the identity function $\mathrm{id}_X : x \mapsto x$ is continuous as a function from $\langle X , \mathcal{O} \rangle$ to $\langle X , \mathcal{O}^\prime \rangle$, then the topology $\mathcal{O}$ is finer than the topology $\mathcal{O}^\prime$.)

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  • $\begingroup$ Any finite open set in $\mathbb{E}^1$ does not belong to the other topology. Also the set $(-\infty, 0]\cup[1,\infty)$ is open in finite complement topology but does not belong to the other. So I don't think there is a finer set here. $\endgroup$ – Mathew George Jun 25 '14 at 15:43
  • $\begingroup$ @MathewGeorge A finite open set in the standard topology is empty. and $(-\infty,0] \cup [1,\infty)$ is not open in the cofinite topology, since the complement, $(0,1)$, is not finite. $\endgroup$ – Daniel Fischer Jun 25 '14 at 15:56
  • $\begingroup$ I understood my mistake. Thanks. $\endgroup$ – Mathew George Jun 25 '14 at 15:58
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$f$ is continuous since $\mathbb R \setminus$ (any finite set) is open in the standard space $\mathbb R$.

Let $F$ be a finite subset of $\mathbb R$, and let {$x_1,...,x_n$} be the increasing enumeration of $F$. Then $f^{-1} (\mathbb R \setminus F)=(-\infty, x_1)\cup (x_1,x_2)\cup ... \cup (x_n,\infty)$ is open.

Edit: Alternatively you could just notice the closed sets in $X$ are the finite sets, which are also closed in the standard $\mathbb R$.

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