According to the (Banach-)Alaoglu theorem, for any Banach space 
the closed unit ball of 
is compact in the weak* topology (the weakest/coarsest topology that makes all evaluation functionals 
continuous).
For example, take 
. The dual space 
contains an isometric copy of 
because 
. The sequence 
, where 
are the standard basis vectors, is contained in the unit sphere of . Should 
have a weak*-convergent subsequence? Maybe it should, but it does not.
Indeed, take any subsequence 
. If necessary, choose a further subsequence so that 
for all 
. Define
where we set 
. Two things to notice here: (1) 
; and (2) at least 2/3 of the coefficients of 
, 
, have the sign 
. Hence, 
when is odd and 
when is even. This shows that does not converge in the weak* topology.
The above does not contradict the Banach-Alaoglu theorem. Since 
is not separable, the weak* topology on the unit ball of its dual is not metrizable. The compactness can be stated in terms of nets instead of sequences: every bounded net in has a convergent subnet. In particular, the sequence has a convergent subnet (which is not a sequence). I personally find subnets a recipe for erroneous arguments. So I prefer to say: the infinite set 
has a cluster point 
; namely, every neighborhood of contains some 
. You can use the reverse inclusion of neighborhoods to define a subnet, but I’d rather not to. Everything we want to know about can be easily proved from the cluster point definition. For example,
where 
stands for the vector with all coordinates 1.
where 
is the shift operator on
