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I know that the largest vector topology on countable dimensional vector space is sequential (i.e. every sequentially closed set is closed). Does it keep for the arbitrary vector space?

In countable dimensional case I can describe structure of largest vector topology (it coincides with the largest locally convex topology), but I don't know anything about uncountable dimensional case.

(Largest vector topology on a vector space is the supremum of set of all vector topologies on it.)

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    $\begingroup$ It would help if you explained more precisely what topology you're speaking about. $\endgroup$
    – Alex Degtyarev
    Commented Mar 23, 2016 at 12:48
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    $\begingroup$ You should keep to a single identity really. Please consider registering. And addressing the earlier comment. $\endgroup$
    – Loïc Teyssier
    Commented Mar 23, 2016 at 13:45
  • $\begingroup$ What do you mean by "vector topology"? $\endgroup$
    – Sergei Akbarov
    Commented Mar 23, 2016 at 18:37
  • $\begingroup$ @AlexDegtyarev Largest vector topology on a vector space is a supremum of set of all vector topologies on it. $\endgroup$
    – red_alert
    Commented Mar 24, 2016 at 5:28
  • $\begingroup$ @SergeiAkbarov The topology is called vector if operations of addition and multiplication are continious. $\endgroup$
    – red_alert
    Commented Mar 24, 2016 at 5:29

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For a linear space $X$ with a Hamel basis $H$ the largest vector topology seems to coincide with the topology of free linear topological space over the discrete space $H$. If this is true, then we can apply known results on the sequentiality of free linear topological spaces, see e.g. http://arxiv.org/pdf/1602.04857 Theorem 10.12.4 of this paper implies that for a discrete space $X$ the free linear topological space $Lin(X)$ over $X$ is sequential iff $Lin(X)$ is a $k$-space iff $X$ is countable.

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  • $\begingroup$ I proved that it's true. Thank you! $\endgroup$
    – red_alert
    Commented Mar 26, 2016 at 6:13
  • $\begingroup$ Then close this question as being answered. By the way, the equivalence of the sequentiality and the k-space property in $Lin(X)$ follows from the metrizablity of compact sets in $Lin(X)$. $\endgroup$
    – Taras Banakh
    Commented Mar 26, 2016 at 10:40

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