# Dilation of Open Set in Normed Vector Space is Open

## Theorem

Let $\Bbb F$ be a subfield of $\C$.

Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.

Let $U \subseteq X$ be an open set.

Let $\lambda \in \Bbb F \setminus \set 0$.

Then:

$\lambda U$ is open.

## Proof

Let:

$v \in \lambda U$

Then:

$\dfrac v \lambda \in U$

Since $U$ is open, there exists $\epsilon > 0$ such that whenever $v' \in X$ and:

$\ds \norm {\frac v \lambda - \frac {v'} \lambda} < \epsilon$

we have $v'/\lambda \in U$.

That is, $v' \in \lambda U$.

So, whenever:

$\ds \norm {v - v'} < \epsilon \cmod \lambda$

we have $v' \in \lambda U$.

Since $v$ was arbitrary:

$\lambda U$ is open.

$\blacksquare$