By Max-Albert Knus

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33. We have ObsDiam(Sn (1); −κ) = Sep(Sn (1); κ/2, κ/2). Proof. 26(1) implies ObsDiam(Sn (1); −κ) ≤ Sep(Sn (1); κ/2, κ/2). We prove the reverse inequality. Let f (x) := dSn (1) (x0 , x), x ∈ Sn (1) be the distance function from a fixed point x0 ∈ Sn (1), where dSn (1) is the geodesic distance function on Sn (1). We see d f∗ σ n sinn−1 r (r) = , π n−1 d L1 t dt 0 sin which implies diam( f∗ σ n ; 1 − κ) = π − 2v−1 (κ/2). 32. 33 yields ObsDiam(X; −κ) ≤ ObsDiam(Sn (1); −κ) = π − 2v−1 (κ/2). The rest is to estimate π − 2v−1 (κ/2) from above.

D the set Let D be a positive real number and N a natural number. Denote by X≤N of mm-isomorphism classes of finite mm-spaces with cardinality ≤ N and diameter D , we prove the following. ≤ D. 26. Let N ≥ 2. If r ∈ / RN WN and for a real number δ > 0, then ( 0,D ] for an element (r, w) ∈ RN × D (Φ(r, w), X≤N−1 ) ≤ 3δ . [ δ ,D ] Proof. We first assume r ∈ / RN . Then, there are two numbers l, m ∈ {1, . . , N} with l < m such that dΦ(r,w) (xl , xm ) = rlm < δ . We consider the map f : {x1 , . .

XN } such that ν({xl }) = 0 and µΦ(r,w) ({xi }) ≤ ν({xi }) for any i with i = l. It is easy to see that dP (µΦ(r,w) , ν) < δ for any metric on {x1 , . . 12 implies D (Φ(r, w), X≤N−1 ) ≤ (Φ(r, w), (Φ(r, w), ν)) < 2δ . This completes the proof of the lemma. 27. For any natural number N and any positive real number D, the set D is compact with respect to the box metric . X≤N D = {∗} is Proof. We prove the proposition by induction on N. It is trivial that X≤1 D compact. We assume the compactness of X≤N−1 for a natural number N ≥ 2.