90 CHAPTER 3. METRIC SPACES

Theorem 3.11.8 Let U be an open set in R. Then there exist countably many dis-joint open sets {(ai,bi)}∞

i=1 such that U = ∪∞i=1 (ai,bi) .

Proof: Let p ∈U and let z ∈Cp, the connected component determined by p. Since Uis open, there exists, δ > 0 such that (z−δ ,z+δ ) ⊆U. It follows from Theorem 3.11.2that (z−δ ,z+δ ) ⊆ Cp. This shows Cp is open. By Theorem 3.11.6, this shows Cp is anopen interval, (a,b) where a,b ∈ [−∞,∞] . There are therefore at most countably many ofthese connected components because each must contain a rational number and the rationalnumbers are countable. Denote by {(ai,bi)}∞

i=1 the set of these connected components. ■

Definition 3.11.9 A set E in a metric space is arcwise connected if for any twopoints, p,q ∈ E, there exists a closed interval, [a,b] and a continuous function, γ : [a,b]→E such that γ (a) = p and γ (b) = q.

An example of an arcwise connected metric space would be any subset of Rn which isthe continuous image of an interval. Arcwise connected is not the same as connected. Awell known example is the following.{(

x,sin1x

): x ∈ (0,1]

}∪{(0,y) : y ∈ [−1,1]} (3.2)

You can verify that this set of points in the normed vector spaceR2 is not arcwise connectedbut is connected.

Lemma 3.11.10 In Rp, B(z,r) is arcwise connected.

Proof: This is easy from the convexity of the set. If x,y ∈ B(z,r) , then let γ (t) =x+ t (y−x) for t ∈ [0,1] .

∥x+ t (y−x)−z∥ = ∥(1− t)(x−z)+ t (y−z)∥≤ (1− t)∥x−z∥+ t ∥y−z∥< (1− t)r+ tr = r

showing γ (t) stays in B(z,r).■

Proposition 3.11.11 If X ̸= /0 is arcwise connected, then it is connected.

Proof: Let p ∈ X . Then by assumption, for any x ∈ X , there is an arc joining p and x.This arc is connected because it is the continuous image of an interval which is connected.Since x is arbitrary, every x is in a connected subset of X which contains p. Hence Cp = Xand so X is connected. ■

Theorem 3.11.12 Let U be an open subset of Rp. Then U is arcwise connected ifand only if U is connected. Also the connected components of an open set are open sets.

Proof: By Proposition 3.11.11 it is only necessary to verify that if U is connected andopen, then U is arcwise connected. Pick p ∈U . Say x ∈U satisfies P if there exists acontinuous function, γ : [a,b]→U such that γ (a) = p and γ (b) = x.

A≡ {x ∈U such that x satisfies P .}