The goal. Let $X$ be a set endowed with Hausdorff topologies $\tau_w$ and $\tau_n$, such that $\tau_w\subseteq\tau_n$. Let $\mathscr{C}$ denote a family of subsets $A\subseteq X$, which satisfies the following properties.

(i) $\mathscr{C}$ is closed under arbitrary intersections and finite unions;
(ii) every $\tau_n$-compact set belongs to $\mathscr{C}$; and
(iii) every $A\in\mathscr{C}$ is $\tau_w$-compact.

I would like to define a new topology $\tau_\mathscr{C}$ on $X$ which satisfies the following properties.

(1) $\tau_w\subseteq\tau_\mathscr{C}\subseteq\tau_n$;
(2) every $\tau_\mathscr{C}$ compact set belongs to $\mathscr{C}$; and
(3) every $A\in\mathscr{C}$ is $\tau_\mathscr{C}$-compact.


The obvious thing to try is to take the intersection $\tau_\cap$ of all topologies $\tau$ satisfying $\tau_w\subseteq\tau\subseteq\tau_n$ and for which every $\tau$-compact set belongs to $\mathscr{C}$. However, it is far from obvious that $\tau_\cap$ would satisfy (2) or (3).

Probably this is not possible in general. However, we could assume that $X$ is a Banach space, $\tau_n$ is the norm topology, and $\tau_w$ is the weak topology. We could also, if necessary, impose some additional assumptions on $\mathscr{C}$.

  • $\begingroup$ Isn't it obvious that $\tau_\cap$ satisfies $(2)$? I mean, it's the intersection of all topologies satisfying $(1)$ and $(2)$... Therefore, there exists such a topology if and only if the finest such that $(3)$ holds is finer than $\tau_\cap$. $\endgroup$ – Ben Dec 15 '14 at 18:25
  • $\begingroup$ Well, that $\tau_\cap$ satisfies (2) may well be true, and it may have an elementary and/or routine proof. However, it is not sufficiently obvious that I can see it! $\endgroup$ – Ben W Dec 15 '14 at 19:05
  • $\begingroup$ It may help if I explicitly defined $\mathscr{C}$, although the definition may be hard to understand for non-functional-analysts. We let $1\leq\xi<\omega_1$ be a countable ordinal and denote by $\mathcal{S}_\xi$ the $\xi$-order Schreier family. We then define $\mathscr{C}$ by letting $A\subseteq X$ belong to $\mathscr{C}$ if and only if every sequence $(x_n)\subseteq A$ admits a subsequence $(x_{n_k})$ which converges weakly to some $x\in A$, and for which given $\epsilon>0$ there is a convex combination $(a_k)$ with support in $\mathcal{S}_\xi$ such that $\|x-\sum a_kx_{n_k}\|<\epsilon$. $\endgroup$ – Ben W Dec 15 '14 at 19:09
  • $\begingroup$ Ah, you're right. It's not obvious at all... Sorry $\endgroup$ – Ben Dec 16 '14 at 8:19
  • $\begingroup$ @Ben It seems I've made at least two mistakes in my tentative approach. First one: I forgot to consider the supremum of the empty family. When such supremum is taken in $I=[\tau_w,\tau_n]$ it gives $\tau_w$, but when taken in $L_n=I\cap (2)$ it does not give $\tau_w$ unless $\mathscr{C}$ is the set of $\tau_w$-compact sets. Hence $L_n$ does not inherit arbitrary suprema from $I$ (this was used to show that $L_n$ has also arbitrary infima, a claim that I must retract now). [continues...] $\endgroup$ – johndoe Dec 20 '14 at 2:15

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