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:

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

if and only if:

$v = \lambda x$ 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:

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

$\blacksquare$