Open Sets in Pseudometric Space

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

Then $\varnothing$ and $A$ are both open in $P$.

Proof
From the definition of open set:
 * $\forall y \in U: \exists \epsilon \in \R_{>0}: B_\epsilon \left({y}\right) \subseteq U$

where $U$ is an open set of $P$.

That is:
 * One cannot 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 $\varnothing$:
 * $\forall y \in U: \exists \epsilon \in \R_{>0}: B_\epsilon \left({y}\right) \subseteq U$ is vacuously true.

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