Dilation of Subset of Vector Space Distributes over Sum

Theorem
Let $\GF \in \set {\R, \C}$.

Let $X$ be a vector space over $\GF$.

Let $A_1, \ldots, A_n \subseteq X$ and $\lambda \in \GF$.

Then:
 * $\ds \lambda \sum_{j \mathop = 1}^n A_j = \sum_{j = 1}^n \paren {\lambda A_j}$

where:
 * $\ds \lambda \paren \ldots$ denotes dilation by $\lambda$
 * $\ds \sum_{j \mathop = 1}^n A_j$ denotes the linear combination of subsets of a vector space.

Proof
Let $x \in X$.

We have:
 * $\ds x \in \lambda \sum_{j \mathop = 1}^n A_j$

there exists $x_j \in A_j$ for each $j \in \set {1, 2, \ldots, n}$ such that:
 * $\ds x = \lambda \sum_{j \mathop = 1}^n x_j$

This is equivalent to:
 * $\ds x = \sum_{j \mathop = 1}^n \lambda x_j \in \sum_{j = 1}^n \paren {\lambda A_j}$ for some $x_j \in A_j$.

Hence we have:
 * $\ds x \in \lambda \sum_{j \mathop = 1}^n A_j$


 * $\ds x \in \sum_{j = 1}^n \paren {\lambda A_j}$
 * $\ds x \in \sum_{j = 1}^n \paren {\lambda A_j}$

Hence we obtain:
 * $\ds \lambda \sum_{j \mathop = 1}^n A_j = \sum_{j = 1}^n \paren {\lambda A_j}$