Unit Sphere is Closed/Normed Vector Space

Theorem

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

Let $\Bbb S := \set {x \in X : \norm x = 1}$ be a unit sphere in $M$.

Then $\Bbb S$ is closed in $M$.

Proof

Let $\map {B_1} 0 = \set {x \in X : \norm x < 1}$ be an open ball.

Let $\map { {B_1}^-} 0 = \set {x \in X : \norm x \le 1}$ be a closed ball.

Then:

$\ds X = \Bbb S \bigcup \relcomp X {\Bbb S}$

where

$\ds \relcomp X {\Bbb S} = \map {B_1} 0 \bigcup \paren {X \setminus \map { {B_1}^-} 0}$

is the relative complement of $\Bbb S$ in $X$.

We have that Closed Ball is Closed in Normed Vector Space.

By definition, $X \setminus \map { {B_1}^-} 0$ is open.

Furthermore, Open Ball is Open Set in Normed Vector Space.

By Union of Open Sets of Normed Vector Space is Open, $\relcomp X {\Bbb S}$ is open.

By definition, $\Bbb S$ is closed.

$\blacksquare$

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.3$: Normed and Banach spaces. Topology of normed spaces