Dilation of Union of Subsets of Vector Space

Theorem
Let $K$ be a field.

Let $X$ be a vector space over $K$.

Let $\sequence {E_\alpha}_{\alpha \mathop \in I}$ be an indexed set of subsets of $X$.

Let $\lambda \in K$.

Then:


 * $\ds \lambda \bigcup_{\alpha \in I} E_\alpha = \bigcup_{\alpha \in I} \paren {\lambda E_\alpha}$

where $\lambda E_\alpha$ denotes the dilation of $E_\alpha$ by $\lambda$.

Proof
We have:


 * $\ds v \in \lambda \bigcup_{\alpha \in I} E_\alpha$




 * $v = \lambda x$ for some $\ds x \in \bigcup_{\alpha \in I} E_\alpha$.

This is equivalent to:


 * there exists some $\alpha \in I$ such that $v = \lambda x$ for some $x \in E_\alpha$.

This is equivalent to:


 * there exists some $\alpha \in I$ such that $v \in \lambda E_\alpha$.

This is finally equivalent to:


 * $\ds v \in \bigcup_{\alpha \in I} \paren {\lambda E_\alpha}$

So by the definition of set equality, we have:


 * $\ds \lambda \bigcup_{\alpha \in I} E_\alpha = \bigcup_{\alpha \in I} \paren {\lambda E_\alpha}$