The span is monotone and idempotent, $\mathrm{span}(S)=S$ exactly when $S$ is a linear subspace, and $\mathrm{span}(S\cup \{0_V\})=\mathrm{span}(S)$

Let $V$ be a vector space over a field $F$ (Vector space over a field) and let $S,T\subseteq V$. Then:

1. Extensive. $S\subseteq \mathrm{span}(S)$.
2. Monotone. If $S\subseteq T$ then $\mathrm{span}(S)\subseteq \mathrm{span}(T)$.
3. Idempotent. $\mathrm{span}(\mathrm{span}(S))=\mathrm{span}(S)$.
4. $\mathrm{span}(S)=S$ if and only if $S$ is a linear subspace of $V$ (Linear subspace of a vector space).
5. $\mathrm{span}(S\cup \{0_V\})=\mathrm{span}(S)$.

• These three properties are exactly what makes the span a closure operator on the subsets of $V$: extensive, monotone and idempotent. The closed sets of that operator are the linear subspaces, which is the content of claim 4. The subgroup generated by a subset (The subgroup $⟨S⟩$ generated by a subset, the cyclic subgroup $⟨g⟩$, and cyclic groups) is the closure operator of the same shape on the subsets of a group.

• Claim 5 says $0_V$ is invisible to the span. Adding or deleting the zero vector changes nothing, since every linear subspace contains it. It does not follow, and is not claimed here, that deleting any other single vector changes nothing: that question is about linear independence and belongs to a later page.

• Claim 2 is used constantly in the form "a bigger set spans more". It is what lets a spanning set be enlarged freely, and, with claim 3, what lets the span be computed in stages: spanning a set, then spanning the result, gains nothing.