Inspired by Does a homeomorphism preserve the open sets?, I wanted to ask the corresponding question: while $X$ may have a topology $\mathcal T$ and a homeomorphism from $(X,\mathcal T)$ to $(X,\mathcal T')$ such that $\mathcal T\neq \mathcal T'$, under what conditions must $\mathcal T= \mathcal T'$? A simple example is the discrete topology; can this be generalized?

  • $\begingroup$ You must be using some idiosyncratic definition of "auto-homeomorphism" but I'm not sure what it is. By an "auto-homeomorphism" of a topological space $(X,\mathcal T)$ do you just mean a permutation of the set $X$?? $\endgroup$
    – user14111
    Apr 13 at 3:00
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    $\begingroup$ You don't mean "auto-homeomorphism", which would require the same topological space (set+topology) as both domain and co-domain. You mean honeomorphism between two topological spaces on the same underlying set. $\endgroup$ Apr 13 at 3:05
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    $\begingroup$ In a comment to the linked question, Izaak van Dongen gave further examples, for instance the cofinite topology, cocountable topology etc. $\endgroup$
    – Ulli
    Apr 13 at 5:09
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    $\begingroup$ In short, you are asking what are "permutation invariant topologies" (Izaak van Dongen's terminology is clearest here). $\endgroup$
    – PatrickR
    Apr 13 at 20:33
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    $\begingroup$ Just rephrasing: "each bijective function from $X \rightarrow X$ is a homeomorphism from $(X,\mathcal T) \rightarrow (X,\mathcal T)$". $\endgroup$
    – Ulli
    Apr 14 at 6:14

1 Answer 1


I'll expand on my comment from the other question. My claim is that every such permutation-invariant topology is the co-$\kappa$ topology for some cardinal $\kappa$. By this I mean: a set is closed if and only if it's equal to $X$ or it has cardinality strictly less than $\kappa$. These spaces are clearly all permutation-invariant. They include the discrete space ($\kappa > |X|$), the indiscrete space ($\kappa = 1$), the cofinite space ($\kappa = \aleph_0$), the cocountable space ($\kappa = \aleph_1$), etc.

So, suppose $(X, \tau)$ is permutation-invariant. I will take some cardinal arithmetic for granted (of course, I'm assuming Choice), and I'll make use of the fact that if we know $U \subseteq X$ is open/closed and we have $V \subseteq X$ with $|U| = |V|$ and $|X \setminus U| = |X \setminus V|$, then $V$ must also be open/closed. A special case of this is that if $|U| = |V| < |X|$, then $V$ must be open/closed as we automatically obtain $|X \setminus U| = |X \setminus V|$.

We split into some cases:

  • First, suppose that $X$ is finite, and the topology is not the indiscrete topology. Then some proper non-empty subset $\emptyset \subsetneq U \subsetneq X$ is open. It follows from this that some singleton set (and hence every singleton set) is open, for example by considering all the permutations that fix some point in $U$ and intersecting all the images of $U$ under these permutations. So the space is discrete. So from now on, assume that $X$ is infinite.
  • Now, suppose that there is some nonempty open subset $\emptyset \subsetneq U \subseteq X$ with $|U| < |X|$. Let $x \in U$. By cardinal arithmetic, we know that $|U \setminus \{x\}| < |X \setminus U|$. So we can always find a subset $V'$ of $X \setminus U$ with $|U \setminus \{x\}| = |V'|$. Letting $V = V' \cup \{x\}$, we have that $|U| = |V|$ and $|X \setminus U| = |X \setminus V|$, so there is some permutation taking $U$ to $V$. Hence $V$ is open, and hence $U \cap V = \{x\}$ is open, so the topology is discrete. So from now on, assume that every nonempty open subset has the same cardinality as $X$.
  • Now let $A \subsetneq X$ be a proper closed subset. Suppose that $|A| = |X|$. We can also assume that $|X \setminus A| = |X|$ by the previous. Hence any subset of cardinality $|X|$ with complement also of cardinality $|X|$ is open. It follows that $A$ is clopen and if $x \in A$, then $A \setminus \{x\}$ is also clopen, and hence $\{x\}$ is open, so the topology was discrete again. So from now on, assume that every proper closed subset has cardinality less than $|X|$.
  • So let $A \subsetneq X$ be a proper closed subset, now assuming that $|A| < |X|$. Let $B \subseteq X$ be a subset with $|B| \le |A|$. We'd like to show that $B$ is also closed. Again, by cardinal arithmetic, we know that there's a cardinal $\nu$ such that $\nu + |B| = |A|$ and $2\nu < |X \setminus B|$ (take $\nu = |A|$ if $A$ is infinite, and $\nu = |A| - |B|$ otherwise). So there must be disjoint subsets $A_1', A_2' \subseteq X \setminus B$ with $|A_1'| = |A_2'| = \nu$. Then let $A_1 = A_1' \cup B$ and $A_2 = A_2' \cup B$. We have $|A_1| = |A_2| = |A|$, so these subsets $A_1$ and $A_2$ must be closed and $A_1 \cap A_2 = B$, so it follows that $B$ is closed. This completes the classification.

At the end, I have used the following fact:

Lemma. Suppose $(X, \tau)$ is a topological space and whenever $A \subsetneq X$ is closed and $B \subseteq X$ has $|B| \le |A|$, it follows that $B$ is closed. Then $\tau$ is a co-$\kappa$ topology for some $\kappa$.

(Proof. Take $\kappa$ = $\sup\{|A|^+ : \text{$A \subsetneq X$ is closed}\}$)


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