# Tiling the Cantor space

Given the Cantor space $2^\mathbb{N}$, and a set of disjoint open sets $D$, are there any non-trivial upper bounds on the number of further open sets needed to complete the tiling of the space?

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If I understand correctly, you are looking to complete $D$ to a cover of $2^{\mathbb{N}}$ by pairwise disjoint open sets.
If this interpretation is correct, then the answer is that you either need 0 or 1 more open sets, or it is impossible. And this is true in general for topological spaces. Suppose that $D$ is a family of pairwise disjoint open subsets of some topological space $X$. Then there are three possibilities:
• If $D$ covers $X$, then trivially we need no more sets.
• If $D$ does not cover $X$, and $X \setminus \bigcup D$ is open, then $D \cup \{ X \setminus \bigcup D \}$ is a covering of $X$ by pairwise disjoint open sets.
• If $X \setminus \bigcup D$ is not open, then it is impossible to complete $D$ to a covering of $X$ by pairwise disjoint open sets. of $X$. (If $D^\prime$ were some completion of $D$ to a covering of $X$ by pairwise disjoint open sets, then $\bigcup ( D^\prime \setminus D )$ is open, however it is easy to show that $\bigcup ( D^\prime \setminus D ) = X \setminus \bigcup D$, contradicting our assumption that $X \setminus \bigcup D$ is not open!)