FALSE: The union of two linear subspaces is a linear subspace

False claim: if $U$ and $W$ are linear subspaces of a vector space $V$ over a field $F$ (Linear subspace of a vector space), then $U \cup W$ is a linear subspace of $V$.

The corresponding statement for intersections is true and is The intersection of a nonempty family of linear subspaces of $V$ is a linear subspace of $V$. For unions it fails, and it already fails in the plane over any field: with $e_0, e_1 \in F^{2}$ the vectors with coordinates $(1_F, 0_F)$ and $(0_F, 1_F)$ (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$), take $U := \operatorname{span}\{e_0\}$ and $W := \operatorname{span}\{e_1\}$ (Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$). Then $e_0 \in U$ and $e_1 \in W$, so both lie in $U \cup W$, while their sum, with coordinates $(1_F, 1_F)$, lies in neither.

• Intersections behave, unions do not. An intersection of linear subspaces is always a linear subspace (The intersection of a nonempty family of linear subspaces of $V$ is a linear subspace of $V$), which is what makes $\operatorname{span}(S)$ definable as the smallest linear subspace containing $S$ (Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$). On the union side there is no corresponding construction, and the repair is to take the sum rather than the union: $U + W$ is a linear subspace, and it is exactly $\operatorname{span}(U \cup W)$ ($\sum_{i<n} U_i = \operatorname{span}\bigl(\bigcup_{i<n} U_i\bigr)$, so the sum is the smallest linear subspace containing every $U_i$).

• The exact condition. For linear subspaces $U, W$ of a vector space $V$, the union $U \cup W$ is a linear subspace if and only if $U \subseteq W$ or $W \subseteq U$. One direction is immediate, the union then being the larger of the two. For the other, suppose neither inclusion holds and choose $u \in U \setminus W$ and $w \in W \setminus U$. If $u + w$ were in $U$ then $w = (u + w) + (-u) \in U$, and if $u + w$ were in $W$ then $u = (u + w) + (-w) \in W$; both contradict the choice, since a linear subspace is closed under addition and under additive inverses. So $u + w \notin U \cup W$ and the union is not closed under addition.

• The witness above is an instance of that criterion, since neither of $U$ and $W$ contains the other: $e_0 \in U$ has coordinate $1_F \ne 0_F$ at index $0$ and so is not of the form $(0_F, \lambda)$, and symmetrically for $e_1$.