Open Sets in Vector Spaces with Equivalent Norms Coincide
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Theorem
Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be normed vector spaces.
Let $U \subseteq X$ be an open set in $M_a$.
Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are equivalent norms, i.e. $\norm {\, \cdot \, }_a \sim \norm {\, \cdot \,}_b$.
Then $U$ is also open in $M_b$.
Proof
By definition of equivalent norms:
- $\exists m,M \in \R_{> 0} : m \le M : \forall x \in X: m \norm x_b \le \norm x_a \le M \norm x_b$
Since $U$ is open in $M_a$:
- $\forall x \in U : \exists \epsilon_a \in \R_{> 0} : \map {B_{\epsilon_a}} x \subseteq U$
where $\map {B_{\epsilon_a}} x$ stands for an open ball, defined as:
- $\map {B_{\epsilon_a}} x := \set {\forall y \in X : \norm {x - y}_a < \epsilon_a}$
Define an open ball $\map {B_{\epsilon_b}} x$ as:
- $\map {B_{\epsilon_b}} x := \set {\forall y \in X : \norm {x - y}_b < \epsilon_b}$
Then we have that:
- $m \norm {x - y}_b \le \norm {x - y}_a < \epsilon_a$
So far, $\epsilon_b$ was unspecified.
Define $\epsilon_b := \dfrac {\epsilon_a} m$.
So:
- $\forall x \in U : y \in \map {B_{\epsilon_a}} x \implies y \in \map {B_{\epsilon_b}} x$
with $\epsilon_a = m \epsilon_b$.
Hence:
- $\forall x \in U : \exists \epsilon_b \in \R_{> 0} : \map {B_{\epsilon_b}} x \subseteq U$
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces