Empty Set is Open in Normed Vector Space
Jump to navigation
Jump to search
Theorem
Let $M = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Then the empty set $\O$ is an open set of $M$.
Proof
By definition, an open set $S \subseteq X$ is one where every point inside it is an element of an open ball contained entirely within that set.
That is, there are no points in $S$ which have an open ball some of whose elements are not in $S$.
As there are no elements in $\O$, the result follows vacuously.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.3$: Normed and Banach spaces. Topology of normed spaces