Euclidean Space is Banach Space

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $m$ be a positive integer.

Then the Euclidean space $\R^m$, along with the Euclidean norm, forms a Banach space over $\R$.


Proof

The Euclidean space $\R^m$ is a vector space over $\R$.

That the norm axioms are satisfied is proven in Euclidean Space is Normed Space.

Then we have Euclidean Space is Complete Metric Space.

The result follows by the definition of a Banach space.

$\blacksquare$