Dilation of Complement of Set in Vector Space

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Theorem

Let $K$ be a field.

Let $X$ be a vector space over $K$.

Let $F \subseteq X$ be a non-empty set.

Let $\lambda \in K$.


Then:

$X \setminus \paren {\lambda F} = \lambda \paren {X \setminus F}$


Proof

It is immediate from the definition of a dilation that if $x \in F$ we have $\lambda x \in \lambda F$.

Conversely, if $\lambda x \in \lambda F$, we have $\lambda x = \lambda y$ for some $y \in F$.

That is, $\lambda \paren {x - y} = 0$.

Since $\lambda \ne 0$, it follows that $x = y$, and so $x \in F$.

So we have $x \in F$ if and only if $\lambda x \in \lambda F$.

So for $x \in X$ we have $x \not \in F$ if and only if $\lambda x \not \in \lambda F$.

That is, $x \in X \setminus F$ if and only if $\lambda x \in X \setminus \paren {\lambda F}$.

It follows that $\lambda x \in \lambda \paren {X \setminus F}$ if and only if $\lambda x \in X \setminus \paren {\lambda F}$.

We can therefore conclude:

$X \setminus \paren {\lambda F} = \lambda \paren {X \setminus F}$

$\blacksquare$