# For which topologies do self-homeomorphisms preserve the collection of open sets?

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?

• 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 $X$?? Apr 13 at 3:00
• 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. Apr 13 at 3:05
• In a comment to the linked question, Izaak van Dongen gave further examples, for instance the cofinite topology, cocountable topology etc.
– Ulli
Apr 13 at 5:09
• In short, you are asking what are "permutation invariant topologies" (Izaak van Dongen's terminology is clearest here). Apr 13 at 20:33
• Just rephrasing: "each bijective function from $X \rightarrow X$ is a homeomorphism from $(X,\mathcal T) \rightarrow (X,\mathcal T)$".
– Ulli
Apr 14 at 6:14

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