Is there a Cantor space that contains the interval?

Question

Cantor spaces are topological spaces which are compact, totally disconnected, and perfect. The prototypical example is the Cantor set equipped with the Euclidean topology, although other such examples exist.

Now consider the closed unit interval $X = [0,1]$. Although it is usually equipped with the Euclidean topology $\tau_{d}$, there are many other topologies that can be chosen for $X$ that can give the resulting topological space counter-intuitive properties.

Is there a topology $\tau$ for the closed unit interval such that $(X,\tau)$ is a Cantor space, and so that $\tau$ is finer than the Euclidean topology ($\tau_d \subseteq \tau$)? If not, is there a straightforward way of showing that such a thing is impossible?

Answer 1

I'm really not sure whether this is what you had in mind, since with this new formulation the answer is (almost trivially) "no": Since $(X, \tau)$ is compact, there is no strictly coarser Hausdorff topology on $X$. Hence $\tau_d = \tau$. Hence $\tau$ is not totally disconnected.

Following the above discussion, didn't you want to know whether there is a finer topology $\tau$, such that $(X, \tau)$ is an open subspace (or any subspace?) of the Cantor set (up to homeomorphism)? Or of a (not necessarily metrizable) Cantor space?

Answer 2

Let $\tau$ be the Euclidean topology on $[0, 1]$ and $\mu = \{U \cup T: U \in \tau, T \subset \mathbb Q\}$. Obviously, $\mu$ is a topology on $[0, 1]$, which is finer than $\tau$. $\{[u, v]: u,v \in \mathbb Q\} \cup \{\{q\}\}: q \in \mathbb Q\}$ is a countable, clopen base of $\mu$. Hence, $\mu$ is metrizable, separable and zero-dimensional. Therefore, it is homeomorphic to a subspace of $2^\omega$, which is homeomorphic to the Cantor set.
Note that this subspace is not open in the Cantor set, since it contains isolated points.
Of course, the analogous construction works with the open interval $(0, 1)$ as well.