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I have a problem understanding the meaning of this question:

Consider a set $M$ endowed with the discrete, the indiscrete or the cofinite topology, under which assumptions on M does this yeld a compact space?

As I understood this, I first endow $M$ with the discrete topology, then with the indiscrete and lastly with the cofinite. But in my opinion with the indiscrete topology we don't need any assumptions on M right? But how do I find assumptions with the other two topologies?

Thank you for your help.

user123234
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  • Have you taken sets of various sizes, given them these topologies, and then thought about whether or not they are compact? – Randall Nov 02 '21 at 13:48
  • Adding to @Randall's comment, why do you think that for the indiscrete topology you don't require any assumptions? If that approach is correct, can we have a similar approach for the other topologies? – Aniruddha Deshmukh Nov 02 '21 at 13:50
  • Now I'm not sure anymore if we don't require any assumptions. I wanted to use the proposition which says that M is compact iff for every collection of closed subsets of M with empty intersections we can find finitly many of them which intersects also trivially, but I don't think this is useful – user123234 Nov 02 '21 at 13:50
  • So do you think its better working with the definition? Because then if I look again at the indiscrete topology, then the only open cover of M is M itself, and then we can't find an open subcover, thus M would never be compact, but this is also strange – user123234 Nov 02 '21 at 13:56
  • No, in that case EVERY open cover is already finite. It is trivially compact. – Randall Nov 02 '21 at 13:59
  • Try this exercise: prove that any space that has only a finite number of open sets must be compact. – Randall Nov 02 '21 at 14:02
  • Ah so I was right that with the indiscrete topology we don't need any assumptions to be compact. – user123234 Nov 02 '21 at 14:03
  • Hmm okey, but then with your exercise I now that M has to be finite with the discrete topology, then it is also compact right? But can I use this fact also for the cofinite topology, since in there a subset is open if the complement is finite – user123234 Nov 02 '21 at 14:04
  • On the first, yes, you have the right idea. In the second, you have to be more careful. You need to look at examples under the cofinite topology: do it with $M={1,2}$ and again with $M = \mathbb{R}$. The argument is a little subtle: in the latter, there are an infinite number of open sets. – Randall Nov 02 '21 at 14:07
  • Okey I will try it, maybe I will write here again if I don't get it. – user123234 Nov 02 '21 at 14:09
  • Is it right that if M is finite then the cofinite topology is equal to the descrete topology, therefore compact. Now we only need to consider the infinite case i.e. $card(M)=\infty$. But how do i handle this case? – user123234 Nov 02 '21 at 14:28
  • You are making good progress. If you have an open cover of an infinite $M$, how many points will a single open set miss? – Randall Nov 02 '21 at 15:57
  • sorry I don't get it. a single open set will miss infinitly many points? – user123234 Nov 02 '21 at 18:45
  • No. In the cofinite topology a single open set can miss only finitely many points. – Randall Nov 03 '21 at 00:26
  • Could you explain why? – user123234 Nov 03 '21 at 07:15
  • Because that's the definition of the topology. – Randall Nov 03 '21 at 12:21

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