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Given a vector space $V$ (possibly infinite dimension) and a subspace $W$, is it always possible to write

$$V = W\oplus \overline{W}$$

For some subspace $\overline{W}$ of $V$. How would one show this? Clearly if there is additional structure, like $V$ has an inner product and is finite dimensional we can set $$\overline{W} := W^\perp = \{ v\in V \ | \ \forall w\in W, <w,v> = 0 \}$$ But can we always decompose any vector space $V$ this way given any subspace $W$? This post shows it can be done for finite dimension $W$: Finite dimension case. Is additional structure such as an inner product required?

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    $\begingroup$ It's always possible, requires axiom of choice (if I remember correctly). $\endgroup$
    – Jakobian
    Jul 9, 2019 at 0:12
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    $\begingroup$ Note that you don't need finite-dimensionality if you have an inner product, $W^{\perp}$ is still well-defined. $\endgroup$ Jul 9, 2019 at 0:13
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    $\begingroup$ Look at this post $\endgroup$
    – Jakobian
    Jul 9, 2019 at 0:15
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    $\begingroup$ What if $V$ is a non-closed subspace in infinite dimensions? $\endgroup$ Jul 9, 2019 at 0:24
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    $\begingroup$ @TedShifrin Non-closed? There's no topology involved $\endgroup$
    – Jakobian
    Jul 9, 2019 at 0:47

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Not an answer, but a suggestion

Following @TedShifrin's comment, you might want to consider, in the space $\Bbb R ^ {\Bbb N}$ of all sequences of real numbers, the subspace $V$ consisting of all sequences that are co-finitely zero (i.e., sequences $a_1, a_2, \ldots$ for which there is some number $k$ with $a_i = 0$ for all $i > k$).

It's certainly not immediately clear to me how you'd find a complement to $V$. One challenge is that the constant sequence $1, 1, 1, \ldots$ is a limit of the sequences $1, 0, 0, 0, \ldots$; $1, 1, 0, 0\ldots$; $1,1,1,0, \ldots$; etc., but is not actually an element of $V$.

For an even more extreme case, consider continuous functions from $[0, \infty)$ to $\Bbb R$, and the subspace consisting of those that are eventually zero (i.e., for which there's a number $M$ such that $x > M$ implies that $f(x) = 0$.

I honestly don't know whether either of these provides a counterexample to the existence of complements, but they're the first place I'd look.

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  • $\begingroup$ Linear complement always exists... it's a known fact. Maybe explanation here will be better for you. To find and to prove it exists is a different question. $\endgroup$
    – Jakobian
    Jul 9, 2019 at 0:49
  • $\begingroup$ Um...that's a proof in the finite-dimensional case. But if "it's a known fact," then I guess I'll just have to believe you, because you made your case so clearly. :(\ $\endgroup$ Jul 9, 2019 at 2:27
  • $\begingroup$ No, it's proof in any case. Both finite and infinite case. It relies on the basis extension theorem which requires axiom of choice. No, basis extension theorem doesn't work only in the case when our space is finite-dimensional. $\endgroup$
    – Jakobian
    Jul 9, 2019 at 10:46

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