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I'm reading Axler "Linear agebra done right" and in Chapter 1 he discusses subspaces and direct sum. My question is, are there subspaces of the infinite-dimensional vector spaces, e.g. a functional Banach space with sup norm $V$ that directly sum to the entire space $$V = U_1\oplus U_2$$ (except for trivial $U_1=\{0\}$, $U_2 = V$).

Bayes
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    Not sure if this is what you have in mind but you can decompose the spaces $L^p(\Omega)$ as $L^p(\Omega_1)\oplus L^p(\Omega_2)$ with $\Omega_1\cup\Omega_2=\Omega$. – Dirk May 06 '13 at 20:40
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    But only if $\Omega_1 \cap \Omega_2$ has measure $0$, no? For then, $P_1 = \chi_{\Omega_1}$ and $P_2 = \chi_{\Omega_2}$, as multiplication operators, are idempotents on $L^p(\Omega)$ with $P_1 + P_2 = 1$ and $P_1 P_2 = P_2 P_1 = 0$, giving rise to your decomposition. – Branimir Ćaćić May 06 '13 at 20:55

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Given a vector space $V$, every subspace has a direct complement. This follows from the fact that if $U$ is a subspace we can take a basis of $U$, then complete it to a basis of $V$.

However, unlike the finite dimensional, as we generally cannot write an explicit basis to an infinite dimensional vector space, we need to use an axiom in mathematics which guarantees the existence of one. The axiom of choice allows us to construct bases like that, and so if we assume it - as one often does in modern mathematics - we can always guarantee that there exists a direct complement to any subspace of every vector space.

It is possible to construct mathematical universes in which there are vector spaces which are not spanned by any finite set, and cannot be decomposed into two disjoint subspaces. In fact, the axiom of choice is equivalent to the assertion that in every vector space, every subspace has a direct complement. So just assuming that the axiom of choice fails assures us that there is a vector space that has a subspace without a direct complement.

So we really need it. But sometimes, some spaces are well-behaved enough that we can prove they have nontrivial decompositions; or something even prove that every subspace has a direct complement; or a particular family of subspaces which have a particular form have direct complements. But this depends a lot on the space itself.

Asaf Karagila
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