Open Ball of Point Inside Open Ball

From ProofWiki
Jump to navigation Jump to search


Let $M = \left({A, d}\right)$ be a metric space.

Let $B_\epsilon \left({x}\right)$ be an open $\epsilon$-ball in $M = \left({A, d}\right)$.

Let $y \in B_\epsilon \left({x}\right)$.


$\exists \delta \in \R: B_\delta \left({y}\right) \subseteq B_\epsilon \left({x}\right)$

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.


Let $\delta = \epsilon - d \left({x, y}\right)$.

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

If $z \in B_\delta \left({y}\right)$, then $d \left({y, z}\right) < \delta$.


$d \left({x, z}\right) \le d \left({x, y}\right) + d \left({y, z}\right) < d \left({x, y}\right) + \delta = \epsilon$

Thus $z \in B_\epsilon \left({x}\right)$.

So $B_\delta \left({y}\right) \subseteq B_\epsilon \left({x}\right)$.


This diagram illustrates the proof in $M = \left({\R^2, d_2}\right)$:


In $\R^2$, we can draw a disc radius $\delta$ whose center is $y$ and which lies entirely within the larger disc whose center is $x$ with radius $\epsilon$.