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Let $E$ be a normed space, and $E'$ its dual space. The weak* topology on $E'$ can be defined by the family of seminorms $\{p_x\,|\, x\in E\}$, where for any $f\in E'$, $p_x(f)=|f(x)|$.

My question is, can this topology be defined by fewer seminorms? For example, if we have a countable dense set $\{x_1, x_2, ...\}\subset E$, does the smaller family $\{p_{x_n}\,|n\geq 1\}$ of seminorms define the same topology on $E'$?

One of the difficulties is that open sets in weak* topology are not bounded in the original $\|\cdot\|$ norm.

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    $\begingroup$ Topology generated by a countable number of semi-norms is metrizable but the weak* topology is not, if the space is infinite dimensional. $\endgroup$ Commented Dec 11, 2021 at 5:41
  • $\begingroup$ @Kavi Rama Murthy Thanks! In the end of the post math.stackexchange.com/q/626599 it was claimed that, for a normed space X and its completion $\hat X$, the dual spaces $X'$ and $(\hat X)'$ are isomorphic and have the same weak* topology. However, the weak* topology on the dual $(\hat X)'$ has more seminorms (coming from $\hat X-X$) than on the dual $X'$. Is that claim true? $\endgroup$
    – Yuval
    Commented Dec 11, 2021 at 6:06
  • $\begingroup$ @Kavi Rama Murthy Thanks! Can you provide some hint, or, where I can find a proof? This is a somehow similar situation, where one wants to check the topology remain the same if some extra seminnorms are added into the family... $\endgroup$
    – Yuval
    Commented Dec 11, 2021 at 6:23
  • $\begingroup$ I am sorry. I am deleting my second comment because it was wrong. I have provided an answer to the question in your comment. $\endgroup$ Commented Dec 11, 2021 at 8:44

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For the question in your comment here is a counter-example:

Let $X$ be the space of all finitely non-zero sequences with the sup norm. Its completion is $c_0$ and the dual is $\ell^{1}$. The sequence $(ne_n)$ tends to $0$ in the weak* topology induced by $X$ but not in the weak* topology induced by $c_0$ (since $x_n \to 0$ does not imply $nx_n \to 0$).

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