Singleton in Normed Vector Space is Closed

Theorem
Let $X$ be a vector space over $\R$ or $\C$.

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

Let $x \in X$ be a singleton.

Then $x$ is closed.

Proof
Note that either $\set x = X$ or $\set x \subset X$.

Suppose, $X = \set x$.

Then:


 * $X \setminus \set x = \empty$

We have that Empty Set is Open in Normed Vector Space.

$\set x$ is a complement of $\empty$ in $X$.

By definition, $x$ is closed.

Suppose, $X \ne \set x$, i.e. $\set x \subset X$.

Define $U := X \setminus \set x$.

Let $y \in U$.

Let $\epsilon := \norm {x - y}$.

Suppose, $\epsilon > 0$.

By Normed Vector Space is Open in Itself:


 * $\forall x \in X : \exists \epsilon \in \R_{>0} : \map {B_\epsilon} x \subseteq X$

Suppose, $z \in \map {B_\epsilon} y$.

Then:


 * $\norm {y - z} < \epsilon$.

We have that:

Hence, $z \ne x$.

Then $z$ is in the complement of $\set x$ in $X$, i.e. $z \in U$.

In other words:


 * $\forall y \in U : \exists \epsilon = \norm {x - y} > 0 : \map {B_\epsilon} x \subseteq U$

So $U$ is open, and $\set x = X \setminus U$ is closed.