Open Ball of Point Inside Open Ball

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Theorem

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

Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball in $M = \struct {A, d}$.

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


Then:

$\exists \delta \in \R: \map {B_\delta} y \subseteq \map {B_\epsilon} x$


That is, for every point in an open $\epsilon$-ball in a metric space, there exists an open $\delta$-ball of that point entirely contained within that open $\epsilon$-ball.


Proof

Let $\delta = \epsilon - \map d {x, y}$.

From the definition of open ball, this is strictly positive, since $y \in \map {B_\epsilon} x$.

If $z \in \map {B_\delta} y$, then $\map d {x, y} < \delta$.

So:

$\map d {x, z} \le \map d {x, y} + \map d {y, z} < \map d {x, y} + \delta = \epsilon$

Thus $z \in \map {B_\epsilon} x$.

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

$\blacksquare$


Illustration

This diagram illustrates the proof of Open Ball of Point Inside Open Ball in $M = \struct {\R^2, d_2}$:

NeighborhoodInNeighborhood.png

In $\R^2$, we can draw a circle of radius $\delta$ whose center is $y$ lying entirely within the larger circle of radius $\epsilon$ whose center is $x$ .


Sources