$X$ is a Toronto space if for every $Y \subseteq X$ such that $|Y|=|X|$ then $Y$ is homeomorphic to $X$.

I am trying to prove that every metrizable Toronto space is discrete. I have the following Ask a Topologist post which contains an overview for the proof. However, it uses concepts from descriptive set theory, in particular something about a Cantor-Bendixson resolution, which I have never heard before nor could find anything about.

Does anyone have any hints/tips on how to prove this? What is a Cantor-Bendixson resolution?

The following is a MathJaxified version of the proof given in the above link:

Let $\DeclareMathOperator{\CL}{cl}X$ be a non-discrete metric Toronto space: it means that if $|X| = \kappa$ and $Y$ is a subset of $X$, $|Y| = \kappa$ then $Y$ is homeomorphic to $X$. Then

  1. There is an isolated point in $X$. (Choose two nonempty disjoint open sets $U$, $V$ in $X$. Then the union of $X \setminus U$ and $X \setminus V$ is $X$, hence at least one of these sets has cardinality $\kappa$: assume it is $X \setminus V$. If $y$ is a point in $U$ then $Y = (X \setminus V) \cup \{y\}$ is homeomorphic to $X$ and has an isolated point.)

  2. The set $S$ of isolated points is dense in $X$. (If $\CL(S)$ has cardinality $\kappa$, then we are done. Otherwise $X \setminus \CL(S)$ has cardinality $\kappa$ and does not contain an isolated point, contrary to 1.)

    It follows that if $X$ is not discrete then $|S| = \lambda < \kappa$, specially the density of $X$ is less than $\kappa$.

  3. $X$ is scattered. (For $Y$ subset $X$ let $S(Y)$ denote the set of isolated points in $Y$. Then put $S_0 = S(X)$, $S_\alpha = S(X \setminus \bigcup \{ S_\beta : \beta < \alpha \})$ for $\alpha < \kappa$. As $|S_\alpha| = \lambda$ for every $\alpha < \kappa$, $S_\alpha$ is dense and open in $X \setminus \bigcup \{ S_\beta : \beta < \alpha \}$. Put $Y = \bigcup \{ S_\beta : \beta < \alpha \}$. Thus we get a Cantor-Bendixson resolution of $Y$.)

Up to this point, we used only that $X$ is Hausdorff. If $X$ is metric then (as the weight and density of a metric space are equal and the cardinality of a scattered space is less or equal to its weight, we get a contradiction: $\kappa = |X| \leq \text{density of }X \leq \lambda < \kappa$.


Here are some ideas form the parts of the proof (which involves just concepts from general topology and a tiny bit of set theory):

  1. Note that $Y = ( X \setminus V ) \cup \{ y \}$ is a subspace of $X$ of cardinality $\kappa$, and is therefore homeomorphic to $X$. Also, $y$ is an isolated point of $Y$ (since $V \cap Y = \{ y \}$), and so it must be that $X$ also has an isolated point.

  2. Note that the set of isolated points of $\CL(S)$ is exactly $S$, and so the isolated points of $\CL(S)$ are dense in $\CL(S)$. If $| \CL(S) | = \kappa$, then $\CL(S)$ is homeomorphic to $X$, and therefore the set of isolated points of $X$ (namely, $S$) is dense in $X$. If $| \CL(S) | < \kappa$, then $Y = X \setminus \CL(S)$ must have cardinality $\kappa$, and is therefore homeomorphic to $X$. Note that $Y$ is an open subspace of $X$, and so every isolated point of $Y$ must also be an isolated point of $X$. But $Y$ contains no isolated points of $X$, meaning that $Y$ has no isolated points. But this contradicts the fact that $Y$ is homeomorphic to $X$ since, by the above, $X$ has isolated points!

    Since $X$ is not discrete, then $X$ has non-isolated points. Since every point of $S$ is isolated in $S$, it cannot be that $|S| = \kappa$ (since otherwise $S$ would be homeomorphic to $X$). Therefore $|S| = \lambda < \kappa$. It follows that $d(X) \leq \lambda$ (where $d(X)$ denotes the density of $X$: the smallest cardinal of a dense subset of $X$).

  3. To see that $X$ is scattered, it is a bit simpler, using the idea from this question of yours. Simply form the sequence $\langle I_\alpha \rangle_{\alpha < \kappa}$ as I did in my answer (note the cut off). By induction of $\alpha$ we can show that $| I_\alpha | = \lambda$, and $| X \setminus \bigcup_{\xi < \alpha} I_\xi | = \kappa$. (The latter means that at each stage $X \setminus \bigcup_{\xi < \alpha} I_\xi$ is homoemorphic to $X$, and so its set of isolated points has cardinality $\lambda$. Taking $Y = \bigcup_{\alpha < \kappa} I_\alpha$, note that $Y$ is scattered, and has cardinality $\kappa$, and is therefore homeomorphic to $X$. This means that $X$ is scattered.

    For each $x \in X$ the height of $x$ in $X$ , $\mathrm{ht}(X,x)$, is the unique $\alpha < \kappa$ such that $x \in I_\alpha$

For the ending I will use the following ideas:

  • $d(X) = w(X)$ for any metric space $X$ (where $w(X)$ denotes the weight of $X$: the least cardinality of a base for $X$); and
  • $|X| \leq w(X)$ for any scattered space $X$.

The first is a relatively basic fact you can find in almost any topology text. The latter stems from the fact given any base $\mathcal{B}$ for $X$ for each $x\in X$ there is a $U_x \in \mathcal{B} $ such that $U_x \subseteq \bigcup_{\alpha \leq \mathrm{ht}(X, x)} I_\alpha$ and $U_x \cap I_{\mathrm{ht}(X, x)}=\{x\} $. This yields an injection $X \to \mathcal{B}$.

So from all of this we have that $$\kappa = |X| \leq w(X) = d(X) = \lambda < \kappa,$$ a contradiction!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.