Dilation of Union of Subsets of Vector Space
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Theorem
Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $\family {E_\alpha}_{\alpha \mathop \in I}$ be an indexed family of subsets of $X$.
Let $\lambda \in K$.
Then:
- $\ds \lambda \bigcup_{\alpha \mathop \in I} E_\alpha = \bigcup_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
where $\lambda E_\alpha$ denotes the dilation of $E_\alpha$ by $\lambda$.
Proof
We have:
- $\ds v \in \lambda \bigcup_{\alpha \mathop \in I} E_\alpha$
- $v = \lambda x$ for some $\ds x \in \bigcup_{\alpha \mathop \in I} E_\alpha$.
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:
- $\ds v \in \bigcup_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
So by the definition of set equality, we have:
- $\ds \lambda \bigcup_{\alpha \mathop \in I} E_\alpha = \bigcup_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$
$\blacksquare$