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 $U$.

$\blacksquare$

Sources

• W.A. Sutherland: Introduction to Metric and Topological Spaces (1975): Example $2.3.10$