# Is there an uncountable, non-discrete, Hausdorff Toronto space?

We call a topological space $X$ a Toronto space if for any subspace $Y \subseteq X$ such that $Y$ and $X$ have the same cardinality it follows that $Y$ is homeomorphic to $X$.

Does anybody know what is known about the following question?:

Is there an uncountable, non-discrete, Hausdorff Toronto space?

It is not hard to show that if $X$ is countable, Hausdorff and Toronto then $X$ has the discrete topology. I have been thinking about the uncountable case for a while and it turns out it is a much harder question.

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never been there... is Toronto really THAT boring? –  Yaakov Baruch May 10 '10 at 21:05
Maybe you should try google first. According to wikipedia this is called the Toronto space problem. Look also in mathscinet at review MR1752101 (around 2000): "The original Toronto problem of Steprāns [op. cit.] asked if there is an uncountable non-discrete Hausdorff space which is homeomorphic to each of its uncountable subspaces. This problem is not yet completely solved although Steprāns mentioned some known results in the folklore." –  fherzig May 10 '10 at 21:49
This at least used to be one of the major open problems in set-theoretic topology. See page 15 of the "Open problems in topology book" (now 20 years old) www1.elsevier.com/homepage/sac/opit/book.pdf I have no idea if there has been any progress in the past 20 years. –  Matthew Morrow May 10 '10 at 21:51
Just took a quick look through Open Problems in Topology II, nothing seems to appear, but this really doesn't mean anything. –  Michael Blackmon Feb 2 '11 at 22:09
Yaakov, I'm biased since I'm a Montréalais, but the answer is yes: Toronto is indeed that boring... :) –  François G. Dorais Feb 3 '11 at 0:10

• hence such a space must have derived length $\omega_1$ and be hereditarily separable, thus must be an $S$-space
• this gives a way to have a model where there are no non-discrete Hausdorff Toronto spaces of size $\aleph_1$: assume $2^{\aleph_0}\neq2^{\aleph_1}$ and note that hereditary separability implies that the space has only $2^{\aleph_0}$ autohomeomorphisms while any Toronto space of size $\lambda$ must have $2^\lambda$ autohomeomorphisms
I have no idea what has been proven since then. It is also mentioned that questions concerning Toronto spaces with larger cardinalities and with stronger separation axioms is still open and gives for example the question "Are all regular (or normal) Toronto spaces of size $\aleph_1$ discrete?"
From the 7th status report: in G. Gruenhage, J.T. Moore, Countable Toronto spaces, Fund. Math. 163 (2) (2000) 143–162, they show that there is an $\omega$-Toronto space, where an $\alpha$-Toronto space is a scattered space of Cantor-Bendixson rank $\alpha$ which is homeomorphic to each of its subspaces of rank $\alpha$. They constructed countable $\alpha$-Toronto spaces for each $\alpha\lt\omega$. Gruenhage also constructed consistent examples of countable $\alpha$-Toronto spaces for each $\alpha<\omega_1$. –  Apollo Feb 4 '11 at 19:56