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If $\{A_n\}$ is a strict inductive sequence of Hausdorff locally convex spaces (meaning each of the connecting maps $\iota_{n,n+1}:A_n \rightarrow A_{n+1}$ is a topological inclusion) with each $A_n$ being sequentially complete.

Then is $\operatorname{colim}_{n \in \mathbb{N}}A_n$ (colimit taken in lctvs) necessarily sequentially complete?

The analogous assertion for completeness is true (cf. Schaefer 6.6). I tried taking a sequence in the colimit, and then trying to show it is indeed in a finite stage, but am unable to do so. Any help will be appreciated.

mathable
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  • I think it's a special case of LF spaces. A sequence converges iff it converges in some $A_n$. On countable limit it should be complete, in particular sequentially complete. – Liding Yao Mar 12 '25 at 18:40
  • @Liding : It doesn't follow from that. You are implicitly using that an inclusion of Frechet spaces is automatically a closed inclusion (which follows because of completeness). This argument isn't valid for sequentially complete spaces. – mathable Mar 12 '25 at 18:48
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    This question is now solved (by counterexamples!) on MathOverflow https://mathoverflow.net/questions/489373/strict-inductive-limits-of-sequentially-complete-spaces/489392?noredirect=1#comment1276984_489392 – Jochen Mar 18 '25 at 12:03

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