Normed Vector Space is Open in Itself
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Theorem
Let $M = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Then the set $X$ is an open set of $M$.
Proof 1
By definition, an open set $S \subseteq A$ is one where every point inside it is an element of an open ball contained entirely within that set.
Let $x \in X$.
An open ball of $x$ in $M$ is by definition a subset of $X$.
Hence the result.
$\blacksquare$
Proof 2
By definition, a subset $S \subseteq X$ is open if:
- $\forall x \in X : \exists \epsilon \in \R_{>0} : \map {B_\epsilon} x \subseteq S$
Let $S = X$.
Aiming for a contradiction, suppose $X$ is not open.
By De Morgan's laws:
- $\exists x \in X : \forall \epsilon \in \R_{>0} : \map {B_\epsilon} x \cap \paren {X \setminus S} \ne \O$
Note that:
- $X \setminus S = X \setminus X = \O$.
By Intersection with Empty Set:
- $\map {B_\epsilon} x \cap \O = \O$
Hence:
- $\exists x \in X : \forall \epsilon \in \R_{>0} : \O \ne \O$.
Since Empty Set is Unique, we have a contradiction.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.3$: Normed and Banach spaces. Topology of normed spaces