Let $X=\operatorname{proj}_nX_n$ be a projective limit of a sequence of complete DF-spaces and let $Y$ be its complemented subspace. Does it follow that $Y=\operatorname{proj}_nY_n$ where every $Y_n$ is complemented in $X_n$?
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$\begingroup$ This is a very general question. I would guess that even for projective limits of Banach spaces this is a difficult question. Do you know the answer in that case? $\endgroup$– Jochen WengenrothCommented Jun 16, 2021 at 15:05
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$\begingroup$ Unfortunately, I don't know the answer for projective limits of Banach spaces but I have analyzed the proof that closed subspaces of PLS-spaces are again PLS. If I understand this proof correctly it shows in fact that - in our setting - I may write $Y=\operatorname{proj}_nY_n$ with each $Y_n$ - a closed subspace of $X_n$. $\endgroup$– KrzysztofCommented Jun 17, 2021 at 6:54
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$\begingroup$ This is correct, the key ingredient is that closed subspaces of LS-spaces are LS. For DF-spaces this is no longer true. $\endgroup$– Jochen WengenrothCommented Jun 17, 2021 at 7:11
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$\begingroup$ Yes, I agree with that. Therefore we only know that $Y$ is a projective limit of a sequence of subspaces without claiming that they are DF. This would become true once we knew that $Y_n$ is complemented in $X_n$ for every $n\in\mathbb{N}$. I was hoping to conclude this from the fact that $Y$ is complemented in $X$. $\endgroup$– KrzysztofCommented Jun 17, 2021 at 7:18
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$\begingroup$ @JochenWengenroth, Krzysztof, is there a particular reason why one is lead to think that this could be true? Is the opposite implications true (projective limit of complemented is complemented) or something similar perhaps? Otherwise I don't really understand (a priori) why this would be so difficult $\endgroup$– PelotaCommented May 31 at 2:20
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