Connected Base for Euclidean Space (2.29)

Walter Rudin Principles of Mathematical Analysis, chapter 2, exercise 29:

Prove that every open set in $ℝ^1$ is the union of an at most countable collection of disjoint segments.

Proof: We shall prove something stronger—namely, that every open set in $ℝ^k$ is the union of an at most countable collection of disjoint open connected spaces, which by Theorem 2.47 implies the proposition for $k=1$. This case could in fact be strengthened to path-connected spaces but is beyond the development of current theory.

Lemma 1 (Subspace Nonconnnectedness): Let $E=A∪B$ for disjoint, nonempty, separated $A,B$. Every subset of $E$ containing a point of $A$ and a point of $B$ is nonconnected. Proof: Let $F⊆E$ be such that $C=F∩A$ and $D=F∩B$ are nonempty. Now, note that for spaces $X,Y$ that$$\overline{X∩Y}=(X∩Y)∪(X∩Y)'⊆(X∩Y)∪(X'∩Y')$$$$=(X∪X')∩(Y∪Y')∩(X∪Y')∩(Y∪X')⊆\overline{X}∩\overline{Y}$$So observe $C∪D=(A∪B)∩F=F$ and $\overline{C}∩D=\overline{A∩F}∩D⊆\overline{A}∩\overline{F}∩D⊆\overline{A}∩B$ is empty, and similarly $C∩\overline{D}$ is empty.

Lemma 2 (Upper Bounds of Chains of Connected Spaces): Let $\{E_i\}$ be a collection of connected spaces totally ordered by inclusion. Then $E=∪E_i$ is connected. Proof: Assume $E=A∪B$ for disjoint, nonempty, separated $A,B$. Then some indexed summand $E_α$ must contain a point of $A$, and similarly $E_β$ for $B$. Then by hypothesis either $E_α⊆E_β$ or $E_β⊆E_α$, in either case one of them being nonconnected by Lemma 1, a contradiction.

Lemma 3 (Union of Nondisjoint Connected Spaces): Let $E$ and $F$ be connected spaces with nonempty intersection. Then $E∪F$ is connected. Proof: Assume $E∪F=A∪B$ for disjoint, nonempty, separated $A,B$. We may assume some $x∈E∩F$ is contained in $A$. As well, not all of $E$ and all of $F$ can be contained in $A$ since $B$ is nonempty, hence either $E$ or $F$ must not be connected by Lemma 1.

Now, let $E⊆ℝ^k$ be an open set. Define a relation $\text{~}$ on $E$ by $a \text{~} b$ if there exists a connected subspace of $E$ containing both $a$ and $b$. This is reflexive since $E$ is an open set admitting a neighborhood of any point, and neighborhoods are convex hence connected (cf. Exercise 2.21 and the example below Definition 2.17); clearly symmetrical; and transitive by Lemma 3.

Choose a set of representatives in $E$ inherited from this equivalence relation, and for any representative observe the collection $C$ of connected subspaces of $E$ containing the representative; by Lemma 2 and Zorn's Lemma there exists a maximal connected subspace under inclusion. This maximal subspace must in fact contain every point in the equivalence class, since we might otherwise construct a larger connected subspace by Lemma 3. Hence $E$ is a union of disjoint open connected spaces, and it remains only to prove that there are a countable number of them. Generally, we will prove that there cannot be a disjoint collection of uncountably many nonempty open sets in $ℝ^k$.

Observe any collection of disjoint nonempty open sets in Euclidean space. Take a point from each set, and choose a neighborhood of each so that they are disjoint. However, as we've seen, within each of these neighborhoods we may find a member of the countable base for $ℝ^k$ (cf. Exercises 2.22-23), implying the collection is at most countable.$~\square$