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In volume 4 of Gelfand and Vilenkin they study countably normed spaces. I assume a definition for a countably normed space was given in an earlier volume, but I do not have access to them and I could not find a definition online. How are these spaces defined and how are their topologies constructed?

Judging by the name I am guessing that we have a countable family of norms $\{\|\cdot\|_n\}$. Does a basis for this topology consist of sets of the form $$N_{i_1, \ldots, i_n, \epsilon} = \{x \mid \|x\|_{i_m} < \epsilon, m \leq n\}?$$

My second question is that on Wikipedia one of the requirements for a topological vector space to be a Frechet space is that it is complete with respect to a family of seminorms. But how is completeness defined for a family of norms or seminorms? Do we require anything more than each Cauchy sequence converging with respect to any seminorm/norm in the family? Must these limits agree?

CBBAM
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    Given a countable family of seminorms $|\cdot|n$, you can construct a pseudo-metric via $d(x,y) = \sum{n} \frac{ \min{ |x-y|_n, 1}}{2^n}$. Once you have this pseudo-metric, you get both the topology and the definition of a Cauchy sequence. – Rhys Steele Jan 02 '24 at 11:16
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    And for reference, this is p16 in Vol.2 (found the archive.org link on this page). – Amateur_Algebraist Jan 02 '24 at 11:26
  • @RhysSteele Thank you. I am aware of that fact but I was interested if one can define it directly from the seminorms. – CBBAM Jan 02 '24 at 22:29
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    for a brief definition of the topology from a collection of seminorms see https://math.stackexchange.com/questions/3510982/doubt-in-understanding-space-mathscr-d-omega/3511753#3511753 in particular the $N_{i_1,\ldots,i_n,\epsilon}$ are not a basis of the topology but a neighborhood basis of the origin. To get a basis for the topology you need all the translates of these sets. In the particular situation at hand the completeness amounts to: for each sequence $(x_n)$ which is Cauchy for each $|\cdot|_m$ (obvious analogous definition as in normed space), there exists $x$... – Abdelmalek Abdesselam Jan 06 '24 at 15:35
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    ...such that $\lim_{n\rightarrow\infty}|x_n-x|_m=0$ for all $m$. – Abdelmalek Abdesselam Jan 06 '24 at 15:35
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    @AbdelmalekAbdesselam Thank you very much! – CBBAM Jan 07 '24 at 01:11

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