3.12. PARTITIONS OF UNITY IN METRIC SPACE 91

If x ∈ A, then Lemma 3.11.10 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 ) . Butthen, a repeat of the above argument shows z ∈ A also. Hence B is open and so if B ̸= /0,then U = B∪A and so U is separated by the two sets B and A contradicting the assumptionthat U is connected. Note that, since B is open, it contains no limit points of A and since Ais open, 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 interiorpoint of Cp. ■

As an application, consider the following corollary.

Corollary 3.11.13 Let f : Ω→ Z be continuous where Ω is a connected nonemptyopen set of a metric space. 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. ■

3.12 Partitions of Unity in Metric SpaceLemma 3.12.1 Let X be a metric space and let S be a nonempty subset of X .

dist(x,S)≡ inf{d (x,z) : z ∈ S}

Then|dist(x,S)−dist(y,S)| ≤ d (x,y) .

Proof: Say dist(x,S)≥ dist(y,S) . Then letting ε > 0 be given, there exists z ∈ S suchthat d (y,z)< dist(y,S)+ ε Then

|dist(x,S)−dist(y,S)|= dist(x,S)−dist(y,S)≤ dist(x,S)− (d (y,z)− ε)