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> Every totally disconnected separable metric space of dimension $n$ homeomorphically embeds into $C\times \mathbb R ^n$.
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Is something like this known? $X$ is *totally disconnected* means that every point in $X$ is equal to the intersection of all clopen sets containing the point. $C$ is the Cantor set. 


It is known that there are totally disconnected spaces of arbitrary dimension. But what about just $n=1$? How might we prove $X$ embeds into $C\times [0,1]$? 



Embedding into $C\times [0,1]$
The answer to this question is negative even in dimension $n=1$.


A suitable counterexample can be constructed as follows.


By a [result of Dranishnikov](http://iopscience.iop.org/article/10.1070/SM1987v057n01ABEH003059), there exists a self-map $f:M\_1\to M\_1$ of the Menger cube $M\_1$ such that for any $y\in M\_1$ the preimage $f^{-1}(y)$ is homeomorphic to $M\_1$. Now take any topological copy $Y\subset M\_1$ of $[0,1]$ and consider the preimage $X:=f^{-1}(Y)$. Fix a countable dense subset $Q\subset Y$ and repeating the construction of the Bernstein set, construct a subset $B\subset f^{-1}(Y\setminus Q)$ such that $f{\restriction}B$ is injective and $B$ intersects each compact set $K\subset X$ that has uncountable image $f(K)$. The zero-dimensionality of $Y\setminus Q$ and the injectivity of the map $f{\restriction}B:B\to Y\setminus Q$ implies that the space $B$ is totally disconnected. 


We claim that $B$ cannot be embedded into the product $C\times[0,1]$ of the Cantor set and the interval. Assuming that there exists an embedding $h:B\to C\times[0,1]$, we can use the Lavrentiev Theorem to extend $e$ to a topological embedding $\bar h:G\to C\times[0,1]$ of some $G\_\delta$-set $G\subset f^{-1}(Y\setminus Q)$ of $X$. The construction fo $B$ ensures that the projection $f(X\setminus G)$ is countable. So, we can find a point $y\in Y\setminus f(X\setminus G)$. For this point the preimage $f^{-1}(y)$ is contained in $G$ and hence admits an embedding into $C\times[0,1]$, which is not possible as $f^{-1}(y)$ is homeomorphic to $M\_1$ (which does not embed into $[0,1]$).




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**Remark 1.** The constructed example is essentially non-Borel. It would be interesting to know what happens for totally disconnected Polish spaces of dimension 1. Do they embed into the product $C\times[0,1]$? 


**Remark 2.** A topological space $X$ is called *almost zero-dimensional* if it has a base of the topology consiting of open sets $U$ whose closures $\bar U$ are intersections of open-and-closed sets. By a result of [Dijsktra and van Mill](http://www.ams.org/journals/proc/2007-135-08/S0002-9939-07-08971-X/S0002-9939-07-08971-X.pdf) each almost zero-dimensional space embeds into the complete Erdos space and the latter space embeds into $C\times[0,1]$.


