Every metrizable Toronto space is discrete.

Question

$X$ 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: $\kappa = |X| \leq \text{density of }X \leq \lambda < \kappa$.

Answer 1

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!