Normed Vector Space is Open in Itself/Proof 1
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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, an open set $S \subseteq A$ is one where every point inside it is an element of an open ball contained entirely within that set.
Let $x \in X$.
An open ball of $x$ in $M$ is by definition a subset of $X$.
Hence the result.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis: Chapter $1$: Normed and Banach spaces