in contradiction with (A.3). Thus, there does not exist a convergent subsequence { π$_{m}$$_{k}$ }$_{k}$$_{≥}$$_{1}$ of { π$_{m}$ }$_{m}$$_{≥}$$_{1}$ , completing the contradiction argument. We now return to the contraposition argument. The existence of a sequence { π$_{m}$ }$_{m}$$_{≥}$$_{1}$ of elements of Π not having a convergent subsequence implies that Π is not sequentially compact and, therefore, that it is not compact. By Lemma A.1, Π is not totally bounded. This completes the proof. Recall the definition (4) of F . The next lemma uses the following definitions. We let ℓ $^{∞}$( F ) denote the metric space of all bounded functions z : F → R , equipped with the supremum norm and, for any r > 0, Π$_{r}$ ≡ { π ∈ Π : R ( π ) ≤ r } . If Π ⋆ is nonempty (for instance, if Π is totally bounded by Lemma A.3) we let, for any π ⋆ ∈ Π ⋆ and s > 0, B$_{Π}$ ( π $^{⋆}$, s ) ≡ { π ∈ Π : ‖ π ⋆ − π ‖ ≤ s } denote the intersection of the radius- s L $^{2}$( P ) ball centered at π ⋆ and the collection Π. Because Π ⋆ ⊂ Π, B$_{Π}$ ( π $^{⋆}$, s ) is nonempty. Lemma A.5. Define g : ℓ $^{∞}$( F ) × (0 , ∞ ) → R as Let z ∈ ℓ $^{∞}$( F ) be ‖·‖ -uniformly continuous and let { ( z$_{m}$, r$_{m}$ ) }$_{m}$$_{≥}$$_{1}$ be a sequence with values in ℓ $^{∞}$( F ) × (0 , ∞ ) such that sup$_{f}$$_{∈F}$ | z$_{m}$ ( f ) − z ( f ) | + | r$_{m}$ | → 0 . If Π is totally bounded, then The following corollary will prove useful. Corollary A.6. Recall the definition of g from Lemma A.5. Let h : ℓ $^{∞}$( F ) × [0 , ∞ ) → R be such that, for all z ∈ ℓ $^{∞}$( F ) , g ( z, r ) ≡ sup π ∈ Π$_{r}$ z ( f$_{π}$ ) − lim sup s ↓ 0 sup π $^{⋆}$∈ Π ⋆ inf π ∈ B$_{Π}$ ( π $^{⋆}$,s $_{)}$z ( f$_{π}$ ) . lim sup m →∞ g ( z$_{m}$, r$_{m}$ ) ≤ 0 . h ( z, r ) ≡ max ( g ( z, r ) , 0) if r > 0 and h ( z, 0) ≡ 0 . 25