Normed Vector Space is Open in Itself/Proof 2

Theorem

Let $M = \struct{X, \norm {\, \cdot \,}}$ be a normed vector space.

Then the set $X$ is an open set of $M$.

Proof

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.

$\blacksquare$