FALSE: the union of two linearly independent subsets of a vector space is linearly independent

FALSE. If $A$ and $C$ are linearly independent subsets of a vector space $V$ over a field $F$ (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i<n} \lambda_i v_i = 0_V$ forces every $\lambda_i = 0_F$, and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent), then $A \cup C$ is linearly independent.

• What survives. Two true statements sit either side of the false one. First, every subset of a linearly independent set is linearly independent (Finite sums re-indexed along an injection, with a zero term deleted, and concatenated; and the closure properties of linear independence: an independent list is injective and never $0_V$, its sublists are independent, a list is independent exactly when it is injective with linearly independent image, and every subset of a linearly independent set is linearly independent, claim 7) — independence is inherited downwards, never upwards. Second, adjoining a vector outside the span preserves independence: if $A$ is independent and $w \notin \operatorname{span}(A)$ then $A \cup \{w\}$ is independent (If $S \subseteq V$ is linearly independent and $w \notin \operatorname{span}(S)$ then $S \cup \{w\}$ is linearly independent and $\operatorname{span}(S) \subsetneq \operatorname{span}(S \cup \{w\})$; and if $w \in \operatorname{span}(S)$ then $\operatorname{span}(S \cup \{w\}) = \operatorname{span}(S)$). The false claim is what remains after the second hypothesis is dropped, and dropping it is exactly what goes wrong above: $d = e_0 + e_1$ lies in $\operatorname{span}(A)$.

• Independence is a property of a set, not of its members one at a time. Every single one of $e_0$, $e_1$, $d$ spans an independent singleton, and every two-element subset of $\{e_0,e_1,d\}$ is even a basis of $F^{2}$ ($\{(1,0), (0,1), (1,1)\}$ spans $F^{2}$ and is linearly dependent, so a spanning set need not be a basis; each of its three two-element subsets is a basis); it is only the three together that fail. So no amount of checking pieces establishes independence of a union.

• A sufficient repair, stated but not proved here. If $A$ and $C$ are linearly independent, disjoint, and $\operatorname{span}(A) \cap \operatorname{span}(C) = \{0_V\}$, then $A \cup C$ is linearly independent; an argument of exactly that shape, carried out by hand for three blocks rather than two, is what proves the independence step of The dimension formula: for finite-dimensional linear subspaces $U$ and $W$ of $V$, the subspaces $U + W$ and $U \cap W$ are finite-dimensional and $\dim_F(U+W) + \dim_F(U \cap W) = \dim_F U + \dim_F W$. This library does not record the general statement as a separate item, and nothing above uses it. The span condition is sufficient, not necessary: taking $C := A$ nonempty gives an independent union while $\operatorname{span}(A) \cap \operatorname{span}(C) = \operatorname{span}(A)$ is as large as it can be.