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Let $X$ be a Banach space the associated dual space is denoted by $X^*$. Take $\tau_1$ and $\tau_2$ two topologies in $X^*$ compatible with the duality $(X^*,X)$, such that $\tau_1\subset \tau_2$.

We suppose that $X^*$ is separable for $\tau_1$. Can we say that $X^*$ is separable for $\tau_2$?

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Sure. If $D$ is $\tau_1$ dense, then the linear span of $D$ is $\tau_2$ dense by the separation theorem, and the rational linear combinations of a set in any TVS is dense in the linear span of the set.

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  • $\begingroup$ Why "by the separation theorem" we have the linear span of $D$ is $\tau_2$ dense? $\endgroup$
    – Karim KHAN
    Commented Jun 10, 2020 at 20:28
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    $\begingroup$ @KarimKHAN since $\tau_2$ is compatible with the duality, if rational span of $D$ was not dense there would be $x\in X\0$ such that $<x,d>=0$. $\endgroup$
    – erz
    Commented Jun 10, 2020 at 21:52

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