72 CHAPTER 2. BASIC TOPOLOGY AND ALGEBRA

Proof: From Theorem 2.10.3, f (E) is a connected subset of R. By Theorem 2.10.6f (E) must be an interval. In particular, it must contain y. This proves the corollary. ■

The following theorem is a very useful description of the open sets in R.

Theorem 2.10.8 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 2.10.2that (z−δ ,z+δ ) ⊆Cp. This shows Cp is open. By Theorem 2.10.6, 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 2.10.9 A set E in Rp is arcwise connected if for any two points, p,q ∈E, there exists a closed interval, [a,b] and a continuous function, γ : [a,b]→ E such thatγ (a) = p and γ (b) = q. The set of points γ ([a,b]) is called an arc, Jordan arc, or a simplecurve.

An example of an arcwise connected space would be any subset of Rp which is thecontinuous image of an interval. Arcwise connected is not the same as connected. A wellknown example is the following.{(

x,sin1x

): x ∈ (0,1]

}∪{(0,y) : y ∈ [−1,1]} (2.7)

You can verify that this set of points in R2 is not arcwise connected but is connected.

Lemma 2.10.10 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 2.10.11 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 2.10.12 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.