Does every vector sub-space has an algebraic complement? 
Does every vector sub-space has an algebraic complement?

Question.
Let $V$ be a vector space over a Field. And let $U\subset V$ be a subspace of $V$.
Now, Can we always guarantee that, there exist a subspace $W$ of $V$ such that,
$$V=U\oplus W$$.
means that, for all element $v\in V$ there exists unique $v_1\in U,v_2\in W$ such that 
$$v=v_1+v_2$$
Work
For finite dimemsional case its is easy by extracting the basis of $U$ to $V$.
But for infinite dimemsional can we have such a property?
Thanks in advance!!!
 A: Let C be the family of vector subspaces of V such that for every W in C, $W\cap U=\{0\}$, C is ordered by the inclusion and a totally ordered subset $(W_i)_{i\in I}$ of C has a sup $\cup_{i\in I}W_i$, Zorn lemma implies the existence of a maximal element W of C, $V=U\oplus W$.
A: Yes, this is true even for infinite-dimensional vector spaces. 
Consider the quotient space $V/U$ and the natural map $\pi\colon V\to V/U$. Pick a basis $B$ for $V/U$. For every $b\in B$, choose some $b'\in\pi^{-1}(b)$. Now, construct a map $\varphi\colon V/U\to V$ given by $\varphi(b)=b'$. Now, check that $\pi\circ \varphi=1$ and $V=\varphi(V/U)\oplus U$.

I'll show that every vector space has a basis using Zorn's lemma. Hence, you can extend any linearly independent set to a basis just like what you did in the finite-dimensional cases.

Theorem: Let $V$ be a vector space (or even a module over a division ring) and $L\subset V$ be a linearly independent set. Then, there is a basis of $V$ containing $L$. In particular, such $L$ always exists, namely $\emptyset$. Therefore, $V$ always has a basis. (Here, a set is linearly independent if every its finite subset is linearly independent.)

Proof: Let $F$ be the set of linearly independent subset of $V$ ordered by inclusion. Clearly, $F$ is nonempty. Let $C_1\subset C_2\subset C_3\subset\cdots$ be a chain of $F$. Let $C=\cup_iC_i$. We need to show that $C\in F$. If $v_1,v_2,\cdots,v_n\in C$, then they are contained in some $C_k$. Hence, they are linearly independent. Now, Zorn's lemma implies that there is a maximal element $B\in F$. We claim that $\mathrm{span}B=V$. Assume to the contrary that $\mathrm{span}B\subsetneq V$ and pick $v\in V\setminus\mathrm{span}B$. Consider $B\cup\{v\}$. Since $B$ is maximal, $B\cup\{v\}$ cannot be linearly independent. Hence, there exist scalars $a_1,\cdots,a_{n+1}$, not all zeros, such that
$$a_1b_1+a_2b_2+\cdots+a_nb_n-a_{n+1}v=0$$
Clearly, $a_{n+1}\neq0$, otherwise $a_1=a_2=\cdots=a_n=0$. Hence,
$$v=\frac{1}{a_{n+1}}(a_1b_1+a_2b_2+\cdots+a_nb_n)$$
showing that $v\in\mathrm{span}{B}$, which is a contradiction. QED
Now, to answer your question, pick a basis $B'$ for $U$ and extend it to a basis $B$ for $V$. Let $W=\mathrm{span}(B\setminus B')$ and check that $V=U\oplus W$.
A: I think this works. Without assuming axiom of choice but assuming V has a basis, any finite dimensional subspace N is contained in some complemented subspace C. Either N=C or N is contained in a complemented subspace C' of C. Continuing in this manner one gets a nested sequence of complemented subspaces all containing N. The intersection of these subspaces must be complemented and must be N.
