Does there exist a set in the plane such that topological dimension 2 with empty interior?

Question

I consider as follows, but i could not proceed it. The topological dimension 2 of a set means that there is a base for the open sets of the set consisting of sets U with topological dimension of boundary of U is 1. My question is: if there exists such a base for the open sets of the set, does it mean that it contains a set homeomorphic to open ball?

Answer 1

If we are talking about Lebesgue covering dimension, there is the Menger-Urysohn theorem:

If $M$ is a subset of $\mathbb{R}^{n}$ with empty interior, then $\dim M\leq n-1$.

In brief, the complement $W=\mathbb{R}^{n}\backslash M$ is dense in $\mathbb{R}^{n}$ and we may find sufficiently small coverings $\omega$ of $\mathbb{R}^{n}$ such that the intersection of any $n+1$ elements of $\omega$ lies in $W$. But then $\omega\cap M$ is a small covering of $M$ such that the intersection of any $n+1$ elements is empty. Thus $\dim M\leq n-1$. So the answer to your question is negative.