Hence, c X (M )=∞, as required. For future reference we record in passing that we obtained the following bound when X =L q and 10, any ball B ⊆X of radius r can be covered by K balls of radius 1 2 r. A metric space X is said to be doubling if it is K -doubling for some K ∈N. The metric space M of Theorem 1.9 can be taken to be doubling. Indeed, fix p>2 and n∈N. As in the proof of Theorem 1.9, write ϑ=1/ min{p, 4}. It was shown in [57] that ψ n,p,ϑ (H Z ) is a O(1)-doubling subset of ℓ p . Let S ⊆ℓ p be the disjoint union of translates in ℓ p of the finite sets {ψ n,p,ϑ (B n )} ∞ n=1 that are sufficiently widely-spaced so as to ensure that S is a doubling subset of ℓ p , and sup n∈N c S (M n )<∞. As in the proof of Theorem 1.9, using Theorem 1.7 we get an embedding φ: S !ℓ 1 satisfying ∥φ(x)−φ(y)∥ ℓ1 ≍ ∥x−y∥ ℓp for all x, y∈S. Thus, φ(S)=M is a doubling subset of ℓ 1 .Since S is doubling, by [62] we can extend φ to a Lipschitz function f : ℓ p !ℓ 1 . If there were Lipschitz mappings g: ℓ p !ℓ 2 and h: g(ℓ p )!ℓ 1 such that f =h¤g, then it would follow that, for all x, y∈S, ∥x−y∥ ℓp ≍ ∥φ(x)−φ(y)∥ ℓ1 = ∥h(g(x))−h(g(y))∥ ℓ1 ≲ ∥g(x)−g(y)∥ ℓ2 ≲ ∥x−y∥ ℓp .Therefore, g¤φ −1 would be a bi-Lipschitz embedding of M into ℓ 2 , which we proved above was impossible. We thus arrive at the following statement.Theorem 1.12. For any 2