Let [[tau].sup.*] = ([x.sup.*], [[lambda].sup.*]) be an

accumulation point of the sequence {[x.sup.k], [[lambda].sup.k])} generated by Algorithm 3.

Thus, {[z.sup.k]} has at least one

accumulation point [z.sup.*] = {[n.sub.*],[x.sup.*]).

By the regularity of I, we obtain that [x.sub.0]is a I-sequentially

accumulation point of E.

Assume that A does not have any

accumulation point in X, i.e., there exist r [member of] (0, 1) such that [B.sub.(M,N)](x, r, t) [intersection] A = [empty set] or [B.sub.(M,N)](x, r, t) [intersection] A = {x} for all x [member of] X and for all t > 0.

A common solution is to establish a satellite

accumulation point and to use the lab pack approach to store wastes.

(ii) any

accumulation point of {[x.sup.k]} is the stationary point of (2).

Each such

accumulation point is the limit of a subsequence {[v.sub.kv]}, for which, by Lemma 3.4, the associated sequence {[[sigma].sub.kv+1]} converges, and we denote its limit by [sigma].

is obviously the centre about which all of the squares can be considered rotating and is hence the

accumulation point of the spiral.

In Theorem 9, we prove that every

accumulation point meets the first KT condition, by which the full sequence convergence is provided in Theorem 10.

Intuitively, one would expect this limit to be the first, lowest,

accumulation point. This is of course not strictly true, because for [lambda] [not equal to] 0, the spectrum is everywhere discrete.

The council is listing all rain

accumulation points in the governorate to minimise damage that can be caused by heavy downpours.