1

Let $M(Compact)$ be the set of finite signed measures on a countable set? (with the topology generated by the sets $\left\{ \mu \in M(Compact) : \left| \int f(x) \mu(dx)- a\right| \leq \delta\right\}$ for all $\delta>0$, $a \in R$ and $f \in C_b(Compact)$ (continuous and bounded). (hence weak-*-topology)

Is then $M(Compact)$ first countable?

mimi
  • 853

1 Answers1

1

No, it is not first countable as it is not metrizable. Some arguments are mentioned here, observing that $C_b(Compact)$ has an uncountable Hamel basis.

Community
  • 1
Vobo
  • 2,955
  • I thought metrizable is only sufficient for first countable and not necessary.... Am I wrong? If not, then you only show that it is not metrizable, but not that its not first countable... – mimi Mar 11 '14 at 15:12
  • In topological vector spaces, first countable and metrizable are equivalent. This follows from the fact that the system of 0-neighbourhoods defines the topology everywhere. – Vobo Mar 11 '14 at 19:17