Span equals finite linear combinations

The span of a set consists exactly of its finite linear combinations
Span equals finite linear combinations

Theorem. Let XX be a vector space and let AXA\subset X. Then the of AA is the set

{i=1mαixi | mN, αiK, xiA}, \left\{\sum_{i=1}^m \alpha_i x_i \ \middle|\ m\in\mathbb{N},\ \alpha_i\in K,\ x_i\in A\right\},

i.e., all finite of elements of AA.

Proof sketch. Let YY be the set of all such finite linear combinations. One checks that YY is a linear subspace and contains AA, hence span(A)Y\operatorname{span}(A)\subset Y by minimality. Conversely, span(A)\operatorname{span}(A) is a subspace containing AA, so it is closed under forming finite linear combinations of elements of AA, giving Yspan(A)Y\subset \operatorname{span}(A).