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Let $X$ be an infinite set and let $C(X)$ denote the collection of connected Hausdorff topologies on $X$. Suppose $N\subseteq C(X)$ has the property that whenever $\tau\neq\sigma \in N$ then $(X,\tau)$ and $(X,\sigma)$ are not homeomorphic. In terms of $|X|$, how large can $|N|$ be at most?

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    $\begingroup$ +1 for this very interesting question. BTW what about if we add "compactness", too.? for example, up to homeomorphism, how many topologies on [0 1] make it a compact connected Hausdorff space? $\endgroup$ – Ali Taghavi Apr 17 '16 at 8:36
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    $\begingroup$ when the cardinality of $X$ is $c$, then the cardinality of $N$ is at least $c$. Because there are uncountable number of open contractible subsets of $\mathbb{R}^{3}$, mutually non homeomorphics. $\endgroup$ – Ali Taghavi Apr 17 '16 at 8:50
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    $\begingroup$ @Jeremy: No, it doesn't. Given a set $X$, there is a well defined set of topologies on $X$, and a well defined set of connected Hausdorff topologies on $X$, and homeomorphism is an equivalence relation. So far, no choice. The OP formulated this using a set of representatives which is the tour de force of choice. I, on the other hand, ask what is the cardinality of the set of equivalence classes, which again involves no choice. But of course it wasn't intentional or anything... :-) $\endgroup$ – Asaf Karagila Apr 17 '16 at 10:35
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    $\begingroup$ @Asaf I don't deny that both questions make sense without choice. It was your statement "So you're asking ...", claiming that they're the same question, that I was teasing you about. $\endgroup$ – Jeremy Rickard Apr 17 '16 at 10:54
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    $\begingroup$ @Jeremy: And how does that work for thus far? :-P $\endgroup$ – Asaf Karagila Apr 17 '16 at 10:58
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Let $\kappa:=|X|$. Then $2^{2^\kappa}$ is an obvious upper bound for the number of topologies on $X$.

Every ultrafilter on $\kappa$ will give you a Hausdorff topological space on $\kappa+1$; these are $2^{2^\kappa}$ many spaces. Some of them might be homeomorphic, but there are only $2^\kappa$ many bijections, so you get $2^{2^\kappa}$ many non-homeomorphic spaces.

Assuming that $\kappa$ is at least continuum, you can add one more point, and copies of the unit interval to each point of your original space; this will now give you a connected (even path-connected) space.

If $\kappa$ is smaller than continuum, then replace the unit interval by some countable connected Hausdorff space.

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    $\begingroup$ So if two spaces are not homeomorphic, after your "suspension" adding a point and lots of intervals, they are still hot homeomorphic? $\endgroup$ – Gerald Edgar Apr 17 '16 at 13:00
  • $\begingroup$ @Gerald: Indeed, if I understand the "suspension" procedure correctly, doing it with a one-point space and a two-point space will yield homeomorphic spaces. $\endgroup$ – Pete L. Clark Apr 17 '16 at 14:14
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    $\begingroup$ @GeraldEdgar That doesn't matter if you do the "up to homeomorphism" bit at the end. Goldstern's argument shows that for any infinite $X$ there is a set with the same cardinality $\kappa$ as $X$ and with $2^{2^\kappa}$ connected Hausdorff topologies. Since the number of permutations of this set is only $2^\kappa$, there must be $2^{2^\kappa}$ homeomorphism classes. $\endgroup$ – Jeremy Rickard Apr 17 '16 at 14:35
  • $\begingroup$ Call the original space $A$, the "strings" (copies of the open unit interval or connected space) $B$, and let $c$ be the point where are strings are glued together. As Jeremy Rickard points out, the very many topologies on $A$ induce very many topologies on $A\cup B\cup \{c\}$, and the number of homeomorphism classes stays "very large" (i.e., $2^{2^\kappa}$). -- However, at least in the case $\kappa\ge 2^{\aleph_0}$, it is clear that each autohomeomorphism must map $B$ onto $B$ (all neighborhoods are $\simeq (0,1)$), and $c$ to $c$, hence $A$ onto $A$. $\endgroup$ – Goldstern Apr 17 '16 at 16:20
  • $\begingroup$ My answer suggests that a more interesting question might have been "how many locally connected Hausdorff spaces...". But I am pretty sure that the answer would stay the same. $\endgroup$ – Goldstern Apr 17 '16 at 16:24

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