Finite Subset of Normed Vector Space is Closed

Theorem

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

Let $F \subseteq X$ be finite.

Then $F$ is closed in $M$.

Proof

Suppose $F$ is empty.

By Empty Set is Closed in Normed Vector Space, $F$ is closed.

Suppose, for some $n \in \N$, that:

$\ds F = \bigcup_{i \mathop = 1}^n \set {x_i}$

We have that Singleton in Normed Vector Space is Closed.

Hence $F$ is a finite union of closed sets.

By Finite Union of Closed Sets is Closed in Normed Vector Space, $F$ 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