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Let us consider $\tau=\{G\subset \mathbb R: \mathbb R\setminus G$ is compact set in ($\mathbb R,\tau_u)\}$, where $\tau_u$ denotes the usual topology on $\mathbb R$. Then $\tau$ is a topology coarser than $\tau_u$. Since $(\mathbb R, \tau_u)$ is connected, so $(\mathbb R, \tau)$ is connected. I tried to prove or disprove that $(\mathbb R, \tau_u)$ is compact. But couldn't get any clue. Any hint will be of great help.

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Suppose that $\mathscr{U}$ is a $\tau$-open cover of $\Bbb R$. Pick any $U_0\in\mathscr{U}$, and let $K=\Bbb R\setminus U_0$; $K$ is $\tau_u$-compact, and $\mathscr{U}\setminus\{U_0\}$ is a $\tau$-open cover of $K$. And $\tau\subseteq\tau_u$, so $\mathscr{U}\setminus\{U_0\}$ is also a $\tau_u$-open cover of $K$.

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  • $\begingroup$ @ B.M. Scott: Since every proper closed subspace of $(\mathbb R,\tau)$ is compact, therefore, so is $(\mathbb R, \tau)$. Is it so? $\endgroup$ – user100943 Oct 16 '13 at 6:20
  • $\begingroup$ @user100943: Yes, though you have to explain why the proper closed subspaces are $\tau$-compact. $\endgroup$ – Brian M. Scott Oct 16 '13 at 6:22
  • $\begingroup$ @ B.M. Scott: From your answer, since $\mathcal U\setminus U_0$ is a $\tau_u$ open cover of $K$ and $K$ is $\tau_u$-compact, therefore, it has got a finite subcover and so, $K$ is compact in $(\mathbb R, \tau)$. $\endgroup$ – user100943 Oct 16 '13 at 6:27
  • $\begingroup$ @user100943: That’s right. $\langle K,\tau_u\rangle$ is compact, and $\tau\subseteq\tau_u$, so automatically $\langle K,\tau\rangle$ is compact as well. $\endgroup$ – Brian M. Scott Oct 16 '13 at 6:30
  • $\begingroup$ @ B.M. Scott: Thanks a lot! $\endgroup$ – user100943 Oct 16 '13 at 6:32
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Hint: Show that any set which is compact in the usual topology remains compact in the $\tau$ topology. (This implies that $\mathbb{R} \setminus U$ is $\tau$-compact for any (nonempty) $\tau_u$-open set $U$.)

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