426 CHAPTER 16. STONE’S THEOREM

If F is not one of these Fi, then gF (x) = φ F (x) = φ F (y) = gF (y) = 0. Thus there is nothingto show for these. It suffices to consider the ones above. Restricting W if necessary, we canassume that for x ∈W,

∑F

gF (x) =n

∑i=1

gFi (x)> δ > 0, gFj (x)< ∆ < ∞, j ≤ n

Then, simplifying the above, and letting x,y ∈W, for each j ≤ n,∣∣∣φ Fj(x)−φ Fj

(y)∣∣∣≤ 1

δ2

∣∣∣∣ gFj (x)∑F gF (y)−gFj (y)∑F gF (y)+gFj (y)∑F gF (y)−gFj (y)∑F gF (x)

∣∣∣∣≤ 1

δ2 ∆∣∣gFj (x)−gFj (y)

∣∣+ 1

δ2 ∆

n

∑i=1|gFi (y)−gFi (x)|

≤ ∆

δ2 d (x,y)+

∆

δ2 nd (x,y) = (n+1)

∆

δ2 d (x,y)

Thus on this set W containing z, all φ F are Lipschitz continuous with Lipschitz constant(n+1) ∆

δ2 .

The functions described above are called a partition of unity subordinate to the opencover S. A useful observation is contained in the following corollary.

Corollary 16.1.2 Let S be a metric space and let S be any open cover of S. Then thereexists a set F, an open refinement of S, and functions {φ F : F ∈ F} such that

φ F : S→ [0,1]

φ F is continuous

φ F (x) equals 0 for all but finitely many F ∈ F

∑{φ F (x) : F ∈ F}= 1 for all x ∈ S.

Each φ F is Lipschitz continuous. If U ∈S and H is a closed subset of U, the partition ofunity can be chosen such that each φ F = 0 on H except for one which equals 1 on H.

Proof: Just change your open cover to consist of U and V \H for each V ∈S. Thenevery function but one equals 0 on H and so exactly one of them equals 1 on H.

16.2 An Extension Theorem, RetractsLemma 16.2.1 Let A be a closed set in a metric space and let xn /∈ A,xn → a0 ∈ A andan ∈ A such that d (an,xn)< 6dist(xn,A) . Then an→ a0.

Proof: By assumption,

d (an,a0) ≤ d (an,xn)+d (xn,a0)< 6dist(xn,A)+d (xn,a0)

≤ 6d (xn,a0)+d (xn,a0) = 7d (xn,a0)