Can you express [0,1] as a countable disjoint union of closed sets, other than the trivial way of doing this?

**Answer**

The answer is no. In fact, as Steve D said, we have a theorem that holds for a wide class of spaces, which includes closed intervals, circles, balls and cubes. It was proved by Sierpiński in 1918 [1]. You can find the proof in the book “General Topology” by Ryszard Engelking, but I’ll post here since it’s not easy to find it online. First a definition: a topological space is called a continuum if it is a compact connected Hausdorff space. The precise statement is the following:

**Theorem (Sierpiński).** If a continuum X has a countable cover \{X_i\}_{i=1}^{\infty} by pairwise disjoint closed subsets, then at most one of the sets X_i is non-empty.

In order to prove this we’ll need the following lemmas:

**Lemma 1.** Let X be a continuum. If F is a non-trivial closed subset of X, then for every component C of F we have that \text{Bd}(F) \cap C is non-empty.

*Proof*. Let x_0 be in C. Since X is Hausdorff compact, quasicomponents coincide with components, so C is the intersection of all open-closed sets in F which contain x_0. Suppose that C is disjoint from \text{Bd}(F). Then, by compactness of \text{Bd}(F), there is one open-closed set A in F containing x_0 and disjoint from \text{Bd}(F). Take an open set U such that A = U \cap F. Thus the equality A \cap \text{Bd}(F) = \emptyset implies that A = U \cap \text{Int}(F), so A is open in X. But A is also closed in X, and contains x_0, so A=X. But then \text{Bd}(F) = \emptyset, which is not possible since F would be non-trivial open-closed in X. \bullet

**Lemma 2.** If a continuum X is covered by pairwise disjoint closed sets X_1, X_2, \ldots of which at least two are non-empty, then for every i there exists a continuum C \subseteq X such that C \cap X_i = \emptyset and at least two sets in the sequence C \cap X_1, C \cap X_2, \ldots are non-empty.

*Proof*. If X_i is empty then we can take C = X; thus we can assume that X_i is non-empty. Take j \ne i such that X_j \ne \emptyset. Since X is Hausdorff compact, there are disjoint open sets U,V \subseteq X satisfying X_i \subseteq U and X_j \subseteq V. Let x be a point of X_j and C the component of x in the subspace \overline{V}. Clearly, C is a continuum, C \cap X_i = \emptyset and C \cap X_j \ne \emptyset. By the previous lemma, C \cap \text{Bd}( \overline{V}) \ne \emptyset and since X_j \subseteq \text{Int}(\overline{V}), there exist a k \ne j such that C \cap X_k \ne \emptyset. \bullet

Now we can prove the theorem:

*Proof*. Assume that at least two of the sets X_i are non-empty. From lemma 2 it follows that there exists a decreasing sequence C_1 \supseteq C_2 \ \supseteq \ldots of continua contained in X such that C_i \cap X_i = \emptyset and C_i \ne \emptyset for i=1,2, \ldots The first part implies that \bigcap_{i=1}^{\infty} C_i = \emptyset and from the second part and compactness of X it follows that \bigcap_{i=1}^{\infty} C_i \ne \emptyset. \bullet

The Hausdorff hypothesis is fundamental. For example, consider X a countable infinite set with the cofinite topology. Then X is compact, connected and a T_1-space. However, we can write X as a disjoint union of countable singletons, which are closed.

[1] Sierpiński, W: *Un théorème sur les continus*, Tôhoku Math. J. **13** (1918), 300–305.

**Attribution***Source : Link , Question Author : Kevin Ventullo , Answer Author : t.b.*