Open Sets in Metric Space

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

Then $$\varnothing$$ and $$M$$ are both open in $$M$$.

Proof
From the definition of open, $$\forall y \in U: \exists \epsilon \left({y}\right) > 0: N_{\epsilon \left({y}\right)} \left({y}\right) \subseteq U$$.

"One can not get out of $$U$$ by moving an arbitrarily small distance from any point in $$U$$."

Take the case where $$U = \varnothing$$.

The empty set $$\varnothing$$ is open by dint of the fact that there are no points $$y$$ in $$\varnothing$$ for which the condition is false.

Thus for $$\varnothing$$, $$\forall y \in U: \exists \epsilon \left({y}\right) > 0: N_{\epsilon \left({y}\right)} \left({y}\right) \subseteq U$$ is vacuously true.

When $$U = M$$, there are no points in $$M$$ to which one could get to by leaving $$U$$, an arbitrarily short distance or no, because there are no points in $$M$$ that are outside of $$U$$.