2.11. COMPLETION OF METRIC SPACES 73

Proof: By Proposition 2.10.11 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 .}

If x ∈ A, then Lemma 2.10.10 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.

Now consider B ≡ U \ A. I claim this is also open. If B is not open, there exists apoint z ∈ B such that every open set containing z is not contained in B. Therefore, lettingB(z,δ ) be such that z∈ B(z,δ )⊆U, there exist points of A contained in B(z,δ ) . But then,a repeat of the above argument shows z ∈ A also. Hence B is open and so if B ̸= /0, thenU = B∪A and so U is separated by the two sets B and A contradicting the assumption thatU is connected. Note that, since B is open, it contains no limit points of A and since A isopen, it contains no limit points of B.

It remains to verify the connected components are open. Let z ∈ Cp where Cp is theconnected component determined by p. Then picking B(z,δ ) ⊆ U, Cp ∪B(z,δ ) is con-nected and contained in U and so it must also be contained in Cp. Thus z is an interior pointof Cp. ■

As an application, consider the following corollary.

Corollary 2.10.13 Let f : Ω→ Z be continuous where Ω is a connected nonemptyopen set in Rp. Then f must be a constant.

Proof: Suppose not. Then it achieves two different values, k and l ̸= k. Then Ω =f−1 (l)∪ f−1 ({m ∈ Z : m ̸= l}) and these are disjoint nonempty open sets which separateΩ. To see they are open, note

f−1 ({m ∈ Z : m ̸= l}) = f−1(∪m ̸=l

(m− 1

6,m+

16

))which is the inverse image of an open set while f−1 (l) = f−1

((l− 1

6 , l +16

))also an open

set. ■

2.11 Completion of Metric SpacesLet (X ,d) be a metric space X ̸= /0. Perhaps this is not a complete metric space. In otherwords, it may be that Cauchy Sequences do not converge. Of course if x ∈ X and if xn = xfor all n then {xn} is a Cauchy sequence and it converges to x.

Lemma 2.11.1 Denote by x a Cauchy sequence x being short for {xn}∞

n=1. Then if x,yare two Cauchy sequences, limn→∞ d (xn,yn) exists.