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Let $X$ be a set of continuum cardinality. The group of permutations of $X$ acts on the set of topologies on $X$.

What can be said about the fixed points of this action? Are manifolds fixed points?

What can be said about the finite orbits?

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  • $\begingroup$ The topology of a manifold is very much not a fixed point; for example, on $\mathbb R$, any two subsets of the cardinality of the continuum will be in bijection, whether they are open intervals, closed intervals, or anything else. $\endgroup$
    – LSpice
    Commented Apr 27, 2021 at 18:52
  • $\begingroup$ For a fixed point, in fact, open-ness depends only on the cardinalitty, so there is a minimum cardinality of a non-empty open set, and the open sets are precisely the empty set and those that have at least that cardinality. $\endgroup$
    – LSpice
    Commented Apr 27, 2021 at 18:53
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    $\begingroup$ But meanwhile, the set of all manifolds is fixed set-wise by the action, since any permutation of the underlying set is a (homeomorphic) manifold. Perhaps that is what the OP meant? The set of instances of any truly topological property will be fixed in this sense. $\endgroup$ Commented Apr 27, 2021 at 18:57
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    $\begingroup$ Ah, that is clearly not true, as @LSpice mentions. (I tried to save you!) $\endgroup$ Commented Apr 27, 2021 at 19:03
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    $\begingroup$ @LSpice Your claim that any two subsets of cardinality continuum will be in bijection is not quite right, since you need the complements also to be equinumerous, in order to have a bijection of the whole space. But it doesn't affect your point, since there will be plenty of instances where this is true. $\endgroup$ Commented Apr 27, 2021 at 19:04

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Let $S(X)$ be the group of permutations of $X$, $X$ infinite. It is a classical consequence of the Baer theorem (Onofri for $X$ countable) that $S(X)$ has no nontrivial finite quotient [actually every nontrivial quotient has cardinal $2^{|X|}$]. Hence every finite orbit of $S(X)$ on $T(X)$, the set of topologies on $X$, is a singleton. [Edit: actually $S(X)$ has no proper subgroup of index $<|X|$ and hence this even applies to orbits of cardinal $<|X|$.]

The orbits of $S(X)$ on $2^X$ are indexed by pairs $(u,v)$ of cardinals such that $\max(u,v)=|X|$, namely $C(u,v)$ the set of subsets of cardinal $u$ with complement of cardinal $v$. Let $W$ be the set of such pairs $(u,v)$; it has an obvious total ordering.

Hence an $S(X)$-invariant subset $\tau$ of $2^X$ is determined by a subset $E_\tau$ of $W$. The condition that $\tau$ is stable under taking arbitrary unions means that $(u,v)\le (u',v')$ and $(0,|X|)\neq (u,v)\in E_\tau$ implies $(u',v')\in E_\tau$. In addition, the condition of being stable under taking finite intersection means that $(u,v)\in E_\tau$, $v=|X|$ implies $(u',v')\in E_\tau$ whenever $(u',v')\le (u,v)$, and $(|X|,n)\in E_\tau$ for $0<n<\omega$ implies $(|X|,n')\in E_\tau$ for all $n'\ge n$.

We deduce that every $S(X)$-invariant topology on $X$ is one of the following

  1. the indiscrete topology $\{\emptyset,X\}$;
  2. for some infinite $\alpha\le |X|$, the topology $\tau_\alpha$ for which open subsets are $\emptyset$ and subsets with complement of cardinal $<\alpha$;
  3. the discrete topology (this is the only Hausdorff one).

Note: this is equivalent to ask what are the topologies on $X$ for which the self-homeomorphism group is the whole group of permutations.

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  • $\begingroup$ Is it really Behr, or Baer? (Also, I think "nontrivial quotient" should be "nontrivial finite quotient", right, as in your comment? For example, you can quotient by the subgroup of permutations that move only finitely many elements.) $\endgroup$
    – LSpice
    Commented Apr 27, 2021 at 22:01
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    $\begingroup$ @LSpice of course I mean "nontrivial finite", thanks for pointing out. $\endgroup$
    – YCor
    Commented Apr 27, 2021 at 22:13
  • $\begingroup$ In "whenever $(u, v) \le (u, v)$", should it be "whenever $(u', v') \le (u, v)$"? $\endgroup$
    – LSpice
    Commented Apr 27, 2021 at 22:27
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    $\begingroup$ @LSpice fixed again, thanks $\endgroup$
    – YCor
    Commented Apr 27, 2021 at 22:28

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