Every linearly independent set extends to a basis

Statement

Let $F$ be a field, let $V$ be a vector space over $F$, and let $L \subseteq V$ be a linearly independent set. Then there is a basis $B$ of $V$ with $L \subseteq B$.

Remarks

For finite-dimensional $V$, add to $L$, one at a time, a vector that is not in the span of the current set. This stops after finitely many steps. In general, the proof uses Zorn's lemma. A stronger form: if $L \subseteq S$, with $L$ independent and $S$ spanning, then a basis $B$ exists with $L \subseteq B \subseteq S$. For example, $\{(1, 1, 0)\}$ extends to the basis $\{(1, 1, 0), (0, 1, 0), (0, 0, 1)\}$ of $\R^3$.

Explanations

L1 human

We can always complete an independent set of vectors to a basis. We keep adding vectors that point in new directions until the set reaches the whole space. This shows that an independent set can never be larger than a basis.