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
We first show that:


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

Let:


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

Then there exists:


 * $\ds x \in \bigcap_{\alpha \mathop \in I} E_\alpha$

such that $v = \lambda x$.

Since $x \in E_\alpha$ for each $\alpha \in I$, we then have $v \in \lambda E_\alpha$ for each $\alpha$.

So:


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

This gives:


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

We now show that:


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

This is clear if $\lambda = 0$.

Now consider the case $\lambda \ne 0$.

Let:


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

Then for each $\alpha \in I$ there exists $x_\alpha \in E_\alpha$ such that $v = \lambda x_\alpha$.

Then, for $\alpha, \beta \in I$, we have:


 * $\lambda x_\alpha = \lambda x_\beta$

Dividing by $\lambda \ne 0$ we have:


 * $x_\alpha = x_\beta$

So there exists $x \in X$ such that:


 * $x_\alpha = x$ for all $\alpha \in I$.

Since $x_\alpha \in E_\alpha$ for each $\alpha \in I$, we then have:


 * $\ds x \in \bigcap_{\alpha \mathop \in I} E_\alpha$

so that:


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

This gives:


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

and hence:


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

by the definition of set equality.