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:

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

if and only if:

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

if and only if:

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

if and only if:

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

if and only if:

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

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.

$\blacksquare$