The intersection of a nonempty family of linear subspaces of $V$ is a linear subspace of $V$

Let $V$ be a vector space over a field $F$ (Vector space over a field) and let $\mathcal{W}$ be a nonempty set of linear subspaces of $V$ (Linear subspace of a vector space). Then

$U={\bigcap }_{W\in \mathcal{W}}W=\{x\in V:x\in W\text{ for every }W\in \mathcal{W}\}$

is a linear subspace of $V$. In particular the intersection of two linear subspaces is a linear subspace.

• The hypothesis that $\mathcal{W}$ is nonempty is load bearing. The intersection of the empty family of subsets of $V$ is not a subset of $V$ by any convention used here, and step 1.1 is where the hypothesis is spent. The same hypothesis appears for the same reason in The intersection of a nonempty family of subgroups of $G$ is a subgroup of $G$.

• This is what makes the span definable. The set of linear subspaces of $V$ containing a given subset $S$ is nonempty, since $V$ itself belongs to it, so its intersection is a linear subspace, and it is by construction the smallest linear subspace containing $S$. That is Linear combination of a finite list, and the span $\mathrm{span}(S)$ as the smallest linear subspace containing $S$, and the pattern is copied from The subgroup $⟨S⟩$ generated by a subset, the cyclic subgroup $⟨g⟩$, and cyclic groups.

• Unions behave quite differently. The union of two linear subspaces is almost never a linear subspace, and the companion examples page records the failure as a false statement.