Dilation of Intersection of Subsets of Vector Space

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:
 * $\ds \lambda \bigcap_{\alpha \mathop \in I} E_\alpha = \bigcap_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$

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

Proof
First, if $\lambda = 0_K$ then we have:
 * $\lambda E_\alpha = \set { {\mathbf 0}_X}$

and:
 * $\ds \lambda \bigcap_{\alpha \mathop \in I} E_\alpha = \set { {\mathbf 0}_X}$

so that:
 * $\ds \bigcap_{\alpha \mathop \in I} \paren {\lambda E_\alpha} = \set { {\mathbf 0}_X} = \lambda \bigcap_{\alpha \mathop \in I} E_\alpha$

Now take $\lambda \ne 0_K$.

Let $v \in X$.

Then, we have:
 * $\ds v \in \lambda \bigcap_{\alpha \mathop \in I} E_\alpha$


 * $\ds \lambda^{-1} v \in \bigcap_{\alpha \mathop \in I} E_\alpha$
 * $\ds \lambda^{-1} v \in \bigcap_{\alpha \mathop \in I} E_\alpha$


 * $\lambda^{-1} v \in E_\alpha$ for each $\alpha \in I$
 * $\lambda^{-1} v \in E_\alpha$ for each $\alpha \in I$


 * $v \in \lambda E_\alpha$ for each $\alpha \in I$
 * $v \in \lambda E_\alpha$ for each $\alpha \in I$


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

So, we have:
 * $\ds \lambda \bigcap_{\alpha \mathop \in I} E_\alpha = \bigcap_{\alpha \mathop \in I} \paren {\lambda E_\alpha}$

in the case $\lambda \ne 0_K$ as well.