Open Ball is Open Set/Metric Space

Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $x \in A$.

Let $\epsilon \in \R_{>0}$.

Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball of $x$ in $M$.


Then $\map {B_\epsilon} x$ is an open set of $M$.


Proof

Let $y \in \map {B_\epsilon} x$.

From Open Ball of Point Inside Open Ball, there exists $\delta \in \R_{>0}$ such that:

$\map {B_\delta} y \subseteq \map {B_\epsilon} x$

The result follows from the definition of open set.

$\blacksquare$


Sources

  • 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{III}$: Metric Spaces: Compactness
  • 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.9$
This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.