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The original question states:

Prove or disprove the following: Let $M,N$ be subspaces of a Banach space $\mathbf{B}$ be such that $M\oplus N = \mathbf{B}$. If $M$ is closed, so is $N$.

The statement is clearly false. For example, $c_0$ is closed but has no closed complement, even though there must exist an $N$ such that $c_0 \oplus N = \ell_\infty$ (That's true in every vector space). $N$ is therefore not closed.

When I talked to my professor about this, he said that he wouldn't accept using the fact that $c_0$ has no complement, as it is not proven in our book and the proof uses advanced tools we don't learn in the course. He also said

You should find a much simpler counterexample. every [Non-finite-dimensional] Banach space has such a counterexample.

However, me and my friends tried finding a counterexample for hours and were unable to. What are we missing?

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  • $\begingroup$ It depends on the meaning of $\oplus$. If it's the direct sum as topological vector spaces, then it's true. $\endgroup$ – Daniel Fischer Jun 14 '18 at 13:58
  • $\begingroup$ Think of linear functionals and their kernels. $\endgroup$ – Daniel Fischer Jun 14 '18 at 14:07
  • $\begingroup$ @DanielFischer What other meanings can it have? $\endgroup$ – Bary12 Jun 14 '18 at 14:07
  • $\begingroup$ Direct sum as vector spaces, ignoring topology. That's the intended meaning here if counterexamples are supposed to exist. $\endgroup$ – Daniel Fischer Jun 14 '18 at 14:08
  • $\begingroup$ Related question $\endgroup$ – A.Γ. Jun 14 '18 at 18:38
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Big hint: If $\lambda$ is an unbounded linear functional on $\bf B$ and $N$ is the nullspace of $\lambda$ then $N$ is not closed, but now if $M=\dots$ then $M$ is closed and ${\bf B}=M\oplus N$.

So you only need to fill in the dots above and then show that if $\bf B$ is infinite-dimensional there exists an unbounded linear functional on $\bf B$ (hint: Hamel basis).

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  • $\begingroup$ Well, to give a counterexample I really only need a single unbounded linear functional in a single Banach space, not to prove that every infinite-dimensional Banach space has one. $\endgroup$ – Bary12 Jun 14 '18 at 14:46

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