250 CHAPTER 13. SOME FUNDAMENTALS∗

Definition 13.9.11 A set E in Rn is arcwise connected if for any two points, p,q ∈E, thereexists a closed interval, [a,b] and a continuous function, γ : [a,b]→ E such that γ (a) = pand γ (b) = q.

An example of an arcwise connected space would be any subset of Rn 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]} (13.2)

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

Lemma 13.9.12 In Rn, 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 13.9.13 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 13.9.14 Let U be an open subset of a Rn. Then U is arcwise connected if andonly if U is connected. Also the connected components of an open set are open sets.

Proof: By Proposition 13.9.13 it is only necessary to verify that if U is connected andopen in the context of this theorem, then U is arcwise connected. Pick p ∈U . Say x ∈Usatisfies P if there exists a continuous function, γ : [a,b]→ U such that γ (a) = p andγ (b) = x.

A≡ {x ∈U such that x satisfies P .}

If x ∈ A, then Lemma 13.9.12 implies B(x,r) ⊆ U is arcwise connected for smallenough r. Thus letting y ∈ B(x,r) , there exist intervals, [a,b] and [c,d] and continuousfunctions having values in U , γ,η such that γ (a) = p,γ (b) =x,η (c) =x, and η (d) = y.Then let γ1 : [a,b+d− c]→U be defined as

γ1 (t)≡

{γ (t) if t ∈ [a,b]η (t + c−b) if t ∈ [b,b+d− c]

Then it is clear that γ1 is a continuous function mapping p to y and showing that B(x,r)⊆A. Therefore, A is open. A ̸= /0 because since U is open there is an open set, B(p,δ )containing p which is contained in U and is arcwise connected.