# Every metrizable Toronto space is discrete.

$X$ 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!