Are complementary subspaces necessarily closed?

I am learning functional analysis by myself and am confused about the definition of complementary subspaces of Banach spaces. In the book I use (Exercise 2.2.1 in "Analysis Now" by Gert Pederson), it seems complementary subspaces are defined to be closed subspaces. I somehow can show that even without closed-ness in the definition, complementary subspaces are closed.

Algebraically, complementary subspaces are the same as they are in the linear algebra $$$$X = X_1 + X_2$$$$ and $$$$X_1 \cap X_2 = \phi$$$$ So any vector $$x\in X$$ has a unique decomposition $$x=P_1 x + P_2 x$$ with $$P_1$$ and $$P_2=1-P_1$$ being the projection operators. Define a new norm $$\lVert x \rVert^{'}= \lVert P_1x \rVert + \lVert P_2x \rVert$$, one can verify it is a norm and dominates the original norm $$\lVert\rVert$$. By the open mapping theorem two norms are equivalent and give the same topology. It's clear from the new norm $$\lVert\rVert^{'}$$ that the projection operators are bounded (continuous). Then the subspaces $$X_1 = P_2^{-1}\{0\}, X_2=P_1^{-1}\{0\}$$ are closed, being the preimage of a one-point (closed) set.

• "By the open mapping theorem..." it doesn't work like that. The theorem is for Banach spaces. Nov 27, 2023 at 9:58
• Every subspace of a vector space has an algebraic complement. But not every subspace of a normed space must have a complement. Thus algebraic complementary subspaces of normed space aren't necessarily closed. So the assumption is important. Nov 27, 2023 at 10:00
• For example, $c_0$ is not complemented in $\ell^\infty$. Nov 27, 2023 at 10:06
• @Jakobian Thanks for the comment. The space $X$ with the new norm $\lVert\rVert^{'}$ must be Banach for me to use the open mapping theorem. If the subspaces are not closed, this is not guaranteed. Nov 27, 2023 at 10:08
• Yes. Your new normed space is Banach iff the subspaces are closed. So you essentially used what you wanted to prove. Nov 27, 2023 at 10:15