The question, which came up in a conversation with my advisor Ola Kwiatkowska, is pretty much in the title:


Let $Z\subseteq[0,1]^3$ be zero-dimensional. Is it possible for $[0,1]^3\setminus Z$ to be embeddable in $\Bbb R^2$?



Can you remove a zero dimensional subspace from a cube and obtain a planar space?
I think the answer is negative in any dimension $n\geq 2$.


**Theorem:**
$\mathbb R^n$ can not be covered by a zero-dimensional set and a set homeomorphic to a subspace of $\mathbb R^{n-1}$.


**Proof:**
Suppose for contradiction that $\mathbb R^n=A\cup B$ where A is zero-dimensional and $i:B\to\mathbb R^{n-1}$ is an embedding. Let $I$ be the interior of $i(B)$ in $\mathbb R^{n-1}$ and let $R=i(B)\setminus I$.
Then $R$ has empty interior in $\mathbb R^{n-1}$ and thus it is known to be of dimension at most $n-2$. Also $\mathbb R^n=A\cup i^{-1}(I)\cup i^{-1}(R)$.
However, $I$ is an open subset of $\mathbb R^{n-1}$, hence its homeomorphic image $i^{-1}(I)$ is not dense in $\mathbb R^n$.
Consequently $A\cup i^{-1}(R)$ is the union of a zero-dimensional space and at most $(n-2)$-dimensional space and covers a non-empty open subset of $\mathbb R^n$, which is a contradiction.


