2 E be affine and let (k) be closed for every k E S. c. / (k) + U for each kE S. c. and affine with T Fix oc >1 If h is a selection for ( § + U 1L+1,P/2(k)r\(h(k) U no ) 0 n( z •0 ) c n,p < P • C§+ U . ) , then for each ke S. c. and affine, so its closure has a continuous affine selection which will also be a selection for ((k) + un+1,13/2) (h(k) + U n, p ) • Now define a sequence { hn 1 n=i CA(S,E) as follows: h 1 (k) E (i(k) + U 1,112 ) — and, if hn is defined, ii n+1 (k)E ((1 (k) + U -n)iN (/ (k) + U -n-1)) ni42 n n+1,2 4:(1(k) Un+1,2 -n-1) • It is clear that (hn ) is Cauchy, and that its limit will have values in If E is a Banach space, the criterion is rather simpler.

Half ‘1, *0;03 3111111• as 0 , b a implies Thus II II bll IIall. (f' - f) A011 3 III' - f 3( E /3) = E . As the norm is additive on S (and hence on -S), we see that II (f' g) A 0 II . As we have seen that 0 it follows that g + (f' - g) A 0 E + (f' - g)A 0 f F, so g belongs to the norm closure of F and the proof is complete. We introduce now a notion that will achieve great importance later. If F is a subset of a compact convex set K, we define the complementary set F' to be the union of all the faces of K that do not meet F.

### Description and theoretical analysis of Planner language by Carl Hewitt

by Richard

4.0