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