A note on the Mackey dual of $C(K)$

Abstract:

Let $K$ be a compact metric space, and let $\tau$ denote the Mackey topology on $M(K)$ with respect to the $\langle C(K),M(K)\rangle$ duality. That is, $\tau$ is the topology of uniform convergence on the weakly compact subsets of $C(K)$. Just as for the weak$^{\ast }$ topology, the dual space of $(M(K),\tau )$ is $C(K)$. However, $\tau$ is very different from weak$^{\ast }$. Indeed, it is obvious that if $\{ {x_n}\}$ is a sequence converging to $x$ in $K$, then $\delta ({x_n})$ converges to $\delta (x)$ in the weak$^{\ast }$ topology, yet Kirk has shown (Pacific J. Math. 45 (1973), 543-554) that $\{ \delta (x)|x \in K\}$ is closed and discrete in the Mackey topology. We obtain a further result along these lines: For each $A \subset K$ set $\Delta A = \{ \delta (x) - \delta (y)|x \ne y,x,y \in A\}$. Let $\mathcal {D}$ denote the totality of all subsets $A$ of $K$ with the property that $0 \in {\overline {\Delta A} ^\tau }$. Then a closed set is in $\mathcal {D}$ iff it is uncountable. Alternatively stated, a closed subset $A$ of $K$ is countable if and only if there is a weakly compact subset $L$ of $C(K)$ such that for every pair $x,\;y \in A,\;x \ne y$, there is an $h \in L$ with $|h(x) - h(y)| \geqslant 1$.