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.
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.
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.