# About countable union

I need to proof that $\mathbb{R}$ can't be written like a countable union of disjoint closed no empty.

If anyone can give me a hint I would really appreciate it.

EDIT: Note that the closed aren't necessarily intervals.

(If I consider the collection $C=\lbrace \mathbb{R} \rbrace$ the problem is trivial).

## 1 Answer

To be clear, we want prove the following: Let $\{X_n\}_{n=1}^{\infty}$ be a collection of non-empty closed subsets of $\mathbb{R}$. Then $\mathbb{R} \neq \cup_{n=1}^{\infty} X_n$. Outline of a proof: By contradiction. Assume that $\mathbb{R} = \cup_{n=1}^{\infty} X_n$. Let $U = \cup_{n=1}^{\infty} X_n^o$ (where $X_n^o$ means the interior of $X_n$). Then $U$ is open. Moreover, $U$ is dense. Indeed, if $\varnothing \neq (a,b) \subset U^c$ then $[a/2,b/2]$ is a non-empty complete metric space which is covered by the countable union of nowhere dense closed sets $\cup_{n=1}^{\infty}\big(X_n \backslash X_n^o\big)$, contradicting the Baire Category Theorem. So $U$ is dense and therefore $U^c$ is nowhere dense and closed. Now, convince yourself that $U^c = \cup_{n=1}^{\infty} \partial X_n$ (where $\partial X_n$ is the boundary of $X_n$). $U^c$ is itself a closed subset and hence a complete metric space (with the induced metric from $\mathbb{R}$). Therefore, we can apply the Baire Category Theorem and infer that since $U^c = \cup_{n=1}^{\infty} \partial X_n$ then there is (at least) one of the sets $\partial X_n$ which is dense in some open subset of $U^c$ (with the induced topology). This says that there is some non-empty open interval $B \subset \mathbb{R}$ for which $B \cap U^c \neq \varnothing$ and $\partial X_n$ is dense in $B \cap U^c$. With $\partial X_n$ closed, then $B \cap U^c \subset \partial X_n$. But this creates a contradiction, since if some $x \in \partial X_n$ is also in $B$, then by the definition of a boundary point there must be some other point $y \in X_m \neq X_n$ such that $y \in B$, which will imply that $\partial X_m \cap B \neq \varnothing$ (why!?) further implying $\partial X_m \cap \partial X_n \neq \varnothing$ from which we can conclude that $X_m$ and $X_n$ are not disjoint.