Dilation of Open Set in Normed Vector Space is Open
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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.
$\blacksquare$