Normed Vector Space is Open in Itself

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