Open Ball of Point Inside Open Ball/Metric Space

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

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

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

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

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

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

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

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

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

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

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

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


 * NeighborhoodInNeighborhood.png

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 $\epsilon$.