A Note on Disjoint Decompositions of [0,1) into Closed Sets

The mechanism

Make the remainder carry the obstruction

A proposed countable partition into closed sets is tested by looking at what remains after its pieces are removed. Closedness in the real line is essential. The argument uses a nonempty closed remainder and Baire category to expose the contradiction.

What becomes elementary: The global partition question is concentrated into a topological obstruction on a carefully chosen remainder.

Keep ambient closedness distinct from relative closedness, and check nonemptiness before applying Baire category.

Explore further constructions. Edition dates, sources and reference chapters identify the scope of each study.

Theorem 1. There does not exist a sequence $(E_n)_{n\geq1}$ of pairwise disjoint closed subsets of $\mathbb R$ such that

[0,1)=\bigcup_{n=1}^{\infty}E_n.

Step 1: Pass to the compact interval [0,1]

Proof. Assume, for contradiction, that such a sequence $(E_n)$ exists. Let

X=[0,1],\qquad F_0=\{1\},\qquad F_n=E_n\quad(n\geq1).

Then $(F_n)_{n\geq0}$ is a sequence of pairwise disjoint closed subsets of $X$ whose union is all of $X$ :

X=\bigcup_{n=0}^{\infty}F_n.

For each $n\geq0$ , let

I_n=\operatorname{int}_X(F_n)

denote the interior of $F_n$ relative to $X$ , and define

B_n=F_n\setminus I_n.

Thus $B_n$ is the part of $F_n$ left over after removing its relative interior. Now set

U=\bigcup_{n=0}^{\infty}I_n,\qquad P=X\setminus U.

Since $U$ is open in $X$ , the set $P$ is closed in $X$ . As $X$ is compact, $P$ is compact and therefore a complete metric space.

The endpoint $1$ lies in no relative interior. Thus $P$ is nonempty. Because the sets $F_n$ are pairwise disjoint and cover $X$ , we have

P=\bigcup_{n=0}^{\infty}B_n,

and each $B_n$ is closed in $P$ .

Step 2: Apply the Baire category theorem

Since $P$ is a complete metric space and is expressed as a countable union of closed subsets, the Baire category theorem implies that at least one of the sets $B_k$ has nonempty interior relative to $P$ . Thus there exist an index $k\geq0$ and a nonempty set $O\subseteq X$ , open in $X$ , such that

\varnothing\neq O\cap P\subseteq B_k.

Choose a point $t\in O\cap P$ . Since $O$ is open in the subspace $X=[0,1]$ , there exists a nondegenerate interval $J\subseteq X$ , open in $X$ , such that

t\in J\subseteq O.

Then

J\cap P\subseteq O\cap P\subseteq B_k.

Step 3: Use disjointness and connectedness

Fix any $m\neq k$ . Since

F_m=I_m\cup B_m

and $B_m\subseteq P$ , the inclusion $J\cap P\subseteq B_k$ implies that

J\cap B_m=\varnothing.

Therefore every point of $J$ lies either in $X\setminus F_m$ or in $I_m$ , so

J\subseteq (X\setminus F_m)\cup I_m.

The sets $X\setminus F_m$ and $I_m$ are disjoint open subsets of $X$ . Since $J$ is connected, it follows that either

J\subseteq X\setminus F_m\qquad\text{or}\qquad J\subseteq I_m.

The second alternative is impossible, because $t\in B_k\subseteq F_k$ while $F_k\cap F_m=\varnothing$ , so $t\notin F_m$ , hence $t\notin I_m$ . Thus

J\subseteq X\setminus F_m\quad\text{for every }m\neq k.

Since the sets $F_n$ partition $X$ , we conclude that

J\subseteq F_k.

But $J$ is open in $X$ , so this implies

J\subseteq\operatorname{int}_X(F_k)=I_k.

In particular, $t\in I_k$ . On the other hand, $t\in J\cap P\subseteq B_k=F_k\setminus I_k$ , so $t\notin I_k$ . This is a contradiction. Hence no such decomposition of $[0,1)$ exists. $\square$

A schematic interval from zero to one, with four thick relative-interior pieces, the remaining closed set P, and a neighborhood J of t whose intersection with P lies in one boundary piece B sub k. — Figure 1. Schematic picture of the proof. The thick subintervals represent relative interiors $I_n$ , while the remaining closed set $P$ consists of the points not lying in any $I_n$ . Baire yields an interval $J$ such that $J\cap P$ lies entirely in a single boundary piece $B_k$ . Connectedness then forces all of $J$ into $F_k$ , contradicting the fact that $t\in B_k\subseteq F_k\setminus I_k$ .