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

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 $$M = \left({\reals^2, d_2}\right)$$:



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 $$\epsilon$$.