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Here's Prob. 2 (a), Sec. 26 in Topology by James R. Munkres, 2nd edition:

Show that in the finite-complement topology on $\mathbb{R}$, every subspace is compact.

I think I can show this.

Now my question is, In this example, is there anything special about $\mathbb{R}$? That is, using the same reasoning, can we not conclude that every set --- finite or infinite --- is compaact under the finite-complement topology?

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    $\begingroup$ There is nothing special about $\Bbb R$; any space with the finite complement topology is hereditarily compact. $\endgroup$ – Brian M. Scott Jul 3 '15 at 7:50
  • $\begingroup$ @BrianM.Scott, it's always so nice to hear from you!! $\endgroup$ – Saaqib Mahmood Jul 3 '15 at 8:15
  • $\begingroup$ @Brian M. Scott, can we give an example of a topological space (other than the discrete topology) no infinite subspace of which is compact? $\endgroup$ – Saaqib Mahmood Jul 5 '15 at 7:37
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    $\begingroup$ One easy example is any uncountable set with the co-countable topology. To get a simple Hausdorff example, start with an uncountable discrete space and add a point whose nbhds are co-countable. $\endgroup$ – Brian M. Scott Jul 5 '15 at 7:58

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