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So I'm trying to solve the following exercise:

If $X$ is a normed space and $Y$ a finite dimensional subspace. Show that $Y$ is complemented in $X.$

I know there's a proof of it using the Hahn-Banach Theorem, however I'm trying to prove it without that. My approach is to use the fact that there exists $Z$ such that $Y\bigoplus Z = X$, thus use the projection $P$ onto $Y$, there is a theorem that says that $ Y$ is topologically complemented if and only if $P$ is bounded. I thought it would be easy to prove that $P$ is bounded but I can't actually prove it.

H_Hassan
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  • The projection is bounded if and only if $Z$ is closed. But if $X$ is infinite-dimensional, then a nontrivial finite-dimensional subspace always has non-closed algebraic complements. So you cannot prove that the projection is bounded without further assumptions. – Daniel Fischer Oct 21 '17 at 16:02
  • So the proof works completely fine if we can find some closed $Z$ that is a complement of $Y$ @DanielFischer – H_Hassan Oct 21 '17 at 16:08
  • Well, yes. But finding a closed complement is more or less the task. – Daniel Fischer Oct 21 '17 at 16:10

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It's fruitless to try to find a proof avoiding the HB theorem, because the statement implies at least this form of the HB theorem:

Theorem: Let $X$ be a NLS, $M$ a subspace and $\phi \in M^*$. Then there exists $\psi \in X^*$ extending $\phi$.

fred goodman
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    OK, perhaps not fruitless, but you would be attempting to find your own proof of HB. – fred goodman Oct 21 '17 at 20:06
  • Could you elaborate a little bit on how the statement implies that version of HB that you state? – Eparoh Oct 16 '23 at 14:31
  • Maybe I wasn't thinking so clearly about this. In any case, the special case of HB where you assume in addition that M is finite dimensional is equivalent to the statement in the original post. – fred goodman Oct 17 '23 at 17:46
  • I see. That special case is in fact evident. – Eparoh Oct 18 '23 at 11:14