1
$\begingroup$

According to this book (see print below) the weak topology (as well as the weak* topology) is first countable. However the weak topology should be first countable only in the finite dimensional case. So, is the book wrong or am I missing something?

The book also says "with regard to convergence we may deal with sequences rather than nets". Is it wrong too?

enter image description here

  • 0
    If you consider sets of the form $N(0;x_1^*,\ldots,x_n^*,1/m)$ for $n,m \in \mathbb N$ you get a neighborhood basis for $0$. – 2017-01-25
  • 0
    @Epsilon this is only countable for a finite dimensional space , because then we have a finite basis for the space of functionals. – 2017-01-25
  • 0
    See http://math.stackexchange.com/q/1417295/4280 as well. Also http://math.stackexchange.com/a/424883/4280 – 2017-01-25
  • 0
    So we do need nets instead of sequences, for this case. – 2017-01-25
  • 0
    conclusion:the book is wrong – 2017-01-25
  • 0
    Okay, true. I had somehow separable spaces in mind. – 2017-01-25
  • 0
    @Epsilon: Even for separable spaces, this is not enough. – 2017-01-25

1 Answers 1

2

I would say your book is plainly wrong. E.g., in every infinite dimensional Banach space equipped with the weak topology, there are sets which are sequentially closed, but not closed, see Is the weak topology sequential on some infinite-dimensional Banach space?. This cannot happen in a first countable space.