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Let $f:M\to S$ be a bijection between two spaces, at least one of which (without loss of generality, $M$) is known to be a manifold. Suppose $f$ preserves and reflects connected sets, meaning $X\subseteq M$ is connected iff $f(X)$ is connected. Must $f$ be a homeomorphism?

It seems to me that the answer is yes, but I'm having trouble formalizing the argument. I think if we prove this for $M=\Bbb R^n$ then we can conclude it for all $n$-manifolds. I believe I have a proof for $n=1$, but I'm having trouble with $n=2$.

Akiva Weinberger
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  • I would be shocked if this were true, but I have no reason to think this. – Randall Oct 25 '22 at 16:48
  • Here's a relevant conjecture: for $n>1$, $A\subseteq\Bbb R^n$ is closed iff for any $p\notin A$, there's a connected set $K\supseteq A$ such that $K\cup{p}$ is disconnected. – Akiva Weinberger Oct 25 '22 at 16:55
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    Related: https://math.stackexchange.com/questions/949168/is-bijection-mapping-connected-sets-to-connected-homeomorphism – Moishe Kohan Oct 25 '22 at 17:00
  • To avoid trivialities you should require that $M$ is connected, or at least that it has only finitely many connected components. (Otherwise, say, $M$ could be $\mathbb{N}$ and $S$ could be $\mathbb{Q}$.) – Eric Wofsey Oct 25 '22 at 23:06
  • @EricWofsey Ah, I had missed that possibility. – Akiva Weinberger Oct 26 '22 at 00:25

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Here is a counterexample for any $M$ of positive dimension. Fix a point $x\in M$ and a sequence of distinct points $(x_n)$ that converges to $x$. Let $T=\{x_n:n\in\mathbb{N}\}$ and $S$ be $M$ with its topology enlarged to make $T$ closed. Explicitly, a subset $U$ of $S$ is open iff is of the form $V\cup (W\setminus T)$ where $V$ and $W$ are open in $M$. I claim that $M$ and $S$ have the same connected sets. Since the topology of $S$ is finer, any $S$-connected set is trivially $M$-connected.

Conversely, suppose $A\subseteq S$ is $S$-disconnected but $M$-connnected. If $x\not\in A$ then this is trivial since the topologies are the same, so we may assume $x\in A$. Let $B\subseteq A$ be a nontrivial $S$-clopen subset, which we may assume contains $x$ (otherwise, take its complement). So $B$ contains $A\cap (U\setminus T)$ for some $M$-open neighborhood $U$ of $x$. If $x_n\in A\cap U$, then since $A$ is $M$-connected, $x_n$ is not isolated in $A$, so $A\cap (U\setminus T)$ contains points that are arbitrarily close to $x_n$. So since $B$ is $S$-closed, $x_n$ must be in $B$, so $B$ actually contains all of $A\cap U$. That is, $B$ actually contains an $M$-open neighborhood of $x$ in $A$, and so $B$ is $M$-open in $A$. Since $x\in B$, $B$ is also $M$-closed in $A$, so $B$ is $M$-clopen in $A$, contradictig the assumption that $A$ was $M$-connected.

(More generally, this construction works if $M$ is any topological space with a subset $T$ that has exactly one accumulation point $x$.)

Eric Wofsey
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  • Interesting! This is a clever construction and a very interesting space. I wonder if $S$ has to be finer (or equal). (In the case that $M$ and $S$ are both manifolds, then, this would imply that they must be the same.) – Akiva Weinberger Oct 26 '22 at 00:33
  • When $M$ and $S$ are both manifolds with finitely many connected components, the topologies must be the same by the argument at https://math.stackexchange.com/questions/949168/is-bijection-mapping-connected-sets-to-connected-homeomorphism. – Eric Wofsey Oct 26 '22 at 00:43
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    Actually, that same argument (using nets instead of sequences) shows that if $M$ is a manifold with finitely many connected components then the topology of $S$ has to be finer. (More generally, it shows this if $M$ is a $T_1$ space such that each point has a neighborhood base of sets whose complements have finitely many connected components.) – Eric Wofsey Oct 26 '22 at 00:54
  • Is there a counterexample where $M$ and $S$ are both manifolds of infinitely many connected components? It seems to me that you can work with each component at a time. – Akiva Weinberger Oct 26 '22 at 03:18
  • Ah, yeah, that would work, since it must map components to components. – Eric Wofsey Oct 26 '22 at 03:24