Every totally disconnected separable metric space of dimension $n$ homeomorphically embeds into $C\times \mathbb R ^n$.
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 $C\times [0,1]$?
We know there is a one-to-one continuous mapping $f:X\to C$ such that the quasi-components of $X$ are the point inverses of $f$. It seems like the $C$-coordinates of the embedding should be determined by $f$, but we need another mapping $g:X\to [0,1]$ determine the second coordinates. At this moment it is unclear how to define $g$. The Erdos space has a norm $\|\;\|$, and the mapping $g$ does something like $$\frac{1}{1+\|x\|}.$$ So maybe just fix any $x'\in X$ and let $$g(x)=\frac{1}{1+d(x,x')}\;\;?$$
But I think this would require that all metric balls around $x'$ have zero-dimensional boundary. With Erdos space this occurs, but it is not clear it always happens with totally disconnected spaces.