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Consider the Eucidean topology over $\mathbb{R}$. Consider the unit interval [0,1]. I want to partition this interval into sets $A$ and $B$ such that

  1. $A \cup B =[0,1]$
  2. Both $A$ and $B$ are closed and both $A$ and $B$ have an empty interior.

It might be the case that such a construction is not possible. In that case, I am interested to know if a closed set with non-empty interior cannot be partitioned into closed sets with empty interior?

For the case of unit interval, I was considering fat Cantor set (https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set) as a possible candidate for set $A$. If set $B$ is the closure of complement of $A$, then $B$ ends up having a non-empty interior.

Math Attack
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ak7019
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2 Answers2

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I am assuming that the use of partition here implies that $A$ and $B$ are intended to be disjoint sets. If so, it is not possible to partition any non-empty topological space into two non-empty closed sets with empty interiors.

If $A$ and $B$ are disjoint and closed, then they must also be open, as they are each other's complement in $X$. But then, the interior of $A$ and $B$ must be themselves; since $X$ is non-empty, one of these must be non-empty as well, concluding the argument.

aghostinthefigures
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No, this is impossible.

Let me assume that $A$ and $B$ are both nonempty and proper subsets of $[0,1]$.

Let me also make the generous assumption that your terminology of a partition of $[0,1]$ into subsets $A$ and $B$ simply means $[0,1] = A \cup B$.

But even under that generous assumption, what you are asking is still impossible.

If $A \subset [0,1]$ is closed in $\mathbb R$ then it is closed in $[0,1]$, i.e. it is relatively closed. Its complement $[0,1]-A$ is relatively open.

It follows that, like every nonempty, relatively open, proper subset of $[0,1]$, the subset $[0,1]-A$ is the disjoint union of a nonempty collection of intervals, each of which is either open, or half-open with its one endpoint at $0$ or $1$ (this, I presume, is the main thing that you were missing).

Since $[0,1]-A \subset B$, it follows that $B$ contains an open interval, and so $B$ has nonempty interior.

Lee Mosher
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