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I have read that the cofinite topology is the coarsest topology on $T_1$, but I have not been able to find any similar statements for the other separation axioms. What are the coarsest topologies for $T_0, T_2, T_3,...$?

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  • $\begingroup$ First you have to check that the intersection of two $T_n$ topologies is a $T_n$ topology (for $n=0,2,3,\ldots$ of course), then you can ask whether or not there is a coarsest. $\endgroup$ – Asaf Karagila Sep 6 '13 at 17:30
  • $\begingroup$ @MJD Since you are probably the user who created (separation-axioms) tag, I thought it might be useful to let you know that there is an ongoing discussion on meta related to this tag. $\endgroup$ – Martin Sleziak Dec 9 '16 at 12:42
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There are probably no such topologies in general. It is the special property of $T_1$ being equivalent to closedness of points which makes this possible for $T_1$. For $T_0$ let's take $κ ∪ \{κ\}$ as system of subsets of $κ$. This system is closed under arbitrary unions and intersections so is a topology on $κ$. It is obviously $T_0$ topology and complements of these sets also forms $T_0$ topology. But any topology coarser than both of these is indiscrete so there is not the coarsest $T_0$ topology on any well-orderable set.

For $T_2$ there is the coarsest topology on any finite set since any Hausdorff topology on finite set is discrete. There is not the coarsest $T_2$ topology on the set of real numbers since in standard topology, non-trivial subset can be open only of is it has cardinality of continuum and its complement has cardinality of continum or is countable. And for these types (if the complement is not finite) there exist some non-open sets which can be by some bijections moved to any other subset of that type. So there are topologies homeomorphic to the standard one such that no non-trivial or non-cofinite subset of $\mathbb{R}$ is open in all of them. So only topology coarser than all of these is cofinite topology which is not $T_2$. In conclusion there is not the coarset topology. Same argument works for any separation axiom satisfied by standard topology on $\mathbb{R}$ like regular, completely regular, normal, completely normal, hereditarily normal, perfectly normal, metrizable, completely metrizable. Similar argument works for any cardinality bigger than continuum. For lesser cardinalities similar argument with $\mathbb{Q}$ could be done.

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  • $\begingroup$ Thanks, @user87690 I will need to ponder this for a bit :) Somehow it seem counter-intuitive that if there is any topology there is not a coarsest one. Perhaps a coarsest equivalence class? $\endgroup$ – ernie Sep 6 '13 at 22:51
  • $\begingroup$ @ernie: It's not any topology. It's a topology such that for any “cardinality type” there is a subset which is not open. There is subset of reals which has cardinality of continuum and its complement is countable and is not open. Since it is not open, it cannot be open in the coarset tpology. But this particular set of that type can be mapped to any such set by some bijection. This bijection also moves the topology so we get homeomorphic topology but the particular set of the type is not open. So in the coarset topology no subset of the type can be open. We do this for all types but cofinite. $\endgroup$ – user87690 Sep 7 '13 at 8:12
  • $\begingroup$ Very interesting. Thank you kindly @user87690 $\endgroup$ – ernie Sep 7 '13 at 15:44

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