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Recall that a Hausdorf topological space $X$ is called compactly generated if any set whose intersections with compacts are compact is closed. Locally compact and first countable spaces are compactly generated.

Let $E$ be a Banach space with the norm $\|\cdot\|$ and the unit ball $B_E$. Let $|||\cdot|||\le \|\cdot\|$ be another norm, and let $\tau$ be a linear (or even locally convex) topology which is stronger than the $|||\cdot|||$-topology, but weaker than the $\|\cdot\|$-topology on $B_E$. Does it follows that $B_E$ with $\tau$ is compactly generated?

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    $\begingroup$ The discrete topology is compactly generated, so you're really asking whether a topology that contains a compactly generated Hausdorff topology must be c. g. $\endgroup$ – Tom Goodwillie Dec 28 '20 at 21:05
  • $\begingroup$ @TomGoodwillie yeah, you are right, and this doesn't seem like it's true, so I'll refocus the question around the specific thing that i need $\endgroup$ – erz Dec 28 '20 at 22:50
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    $\begingroup$ Given any space $(X,\tau)$ it's possible to refine $\tau$ to a compactly generated topology with the same compact sets as $\tau$, so every topology is contained in a very close but compactly generated one without having to go all the way to the discrete one. Going from $\tau_1$ cg to $\tau_2$ cg seems very unlikely to me even though I don't have a counterexample $\endgroup$ – Alessandro Codenotti Dec 28 '20 at 22:57
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Let $\tau$ be the weak topology on the Banach space $\ell_1$. It is known that each weakly convergent sequence in $\ell_1$ is norm convergent (i.e., $\ell_1$ has the Shur property). This property implies that $\tau$ is not compactly generated (otherwise it would be equal to the norm topology). Now consider the norm $$|||(x_n)_{n\in\omega}|||=\sum_{n=0}^\infty\frac{|x_n|}{2^n}$$and observe that the topology $\tau$ is stronger that the topology generated by the norm $|||\cdot|||$ on the unit ball of $\ell_1$.

If we want the topology $\tau$ to be stronger that the topology generated by the norm $|||\cdot|||$ we can replace $\tau$ by the supremum of the weak topology and the topology generated by the norm $|||\cdot|||$. This modified topology still will not be compactly generated (because of the same Shur property).

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