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