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  "body": "\n## Proof\n\nWe extend a basis of $U$ to a basis of $V$ and take the span of the new vectors.\n\n1. By [[every vector space has a basis|00MS9VMQ9Z]], applied to $U$, there is a [[basis|0KCQ9Y3QKA]] $L$ of $U$. We view $L$ as a set of vectors of $U$.\n2. The set $L$ is [[linearly independent|0CRMGPAH83]] in $U$, hence also in $V$, since linear relations are the same in both spaces.\n3. By [[extension to a basis|0M4SXE5WW3]], there is a basis $B$ of $V$ with $L \\subseteq B$.\n4. Let $M = B \\setminus L$, and let $W = \\operatorname{span}(M)$ be its [[linear span|0DTY8B9HMS]].\n5. $W$ is a [[subspace|00YC1D652K]] of $V$. It contains $0$, the empty combination. A sum of two finite linear combinations of $M$, or a scalar multiple of one, is again such a combination.\n6. We show $U + W = V$, with the [[sum of subspaces|0JMTY8MF54]]. Let $v \\in V$. Since $B$ spans $V$, $v$ is a finite linear combination of vectors of $B$.\n7. Group the terms of this combination: those with vectors in $L$ give a vector $u \\in U$, and those with vectors in $M$ give a vector $w \\in W$. Then $v = u + w$.\n8. So $V \\subseteq U + W$. Also $U + W \\subseteq V$, since $U$ and $W$ are subspaces of $V$. Hence $U + W = V$.\n9. We show $U \\cap W = \\{0\\}$. Let $x \\in U \\cap W$.\n10. Since $L$ spans $U$, $x = a_1 \\ell_1 + \\cdots + a_p \\ell_p$ for distinct $\\ell_1, \\dots, \\ell_p \\in L$ and scalars $a_i \\in F$.\n11. Since $x \\in W$, $x = c_1 m_1 + \\cdots + c_q m_q$ for distinct $m_1, \\dots, m_q \\in M$ and scalars $c_k \\in F$.\n12. Subtracting gives $a_1 \\ell_1 + \\cdots + a_p \\ell_p - c_1 m_1 - \\cdots - c_q m_q = 0$.\n13. The sets $L$ and $M$ are disjoint, so the vectors $\\ell_1, \\dots, \\ell_p, m_1, \\dots, m_q$ are distinct elements of $B$.\n14. The basis $B$ is linearly independent, so all $a_i$ and $c_k$ are $0$. Hence $x = 0$.\n15. By steps 8 and 14, $U + W = V$ and $U \\cap W = \\{0\\}$. So $W$ is a [[complement|0M83V17KHX]] of $U$, and $V = U \\oplus W$.\n\n## Idea\n\nA basis of $U$ can be enlarged to a basis of the whole space. The vectors added in this process span a subspace that fills the missing directions and meets $U$ only in $0$. In infinite dimension the enlargement step rests on Zorn's lemma.\n",
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