Neighborhoods: Interior

Question

The neighborhood filters satisfy: $$\forall N\in\mathcal{N}(x):\qquad x\in N$$ $$\forall N\in\mathcal{N}(x)\exists M_0\in\mathcal{N}(x):\qquad N\in\mathcal{N}(m)\text{ for all }m\in M_0$$ Define the interior: $$A^\circ:=\{z:A\in\mathcal{N}(z)\}$$ Prove that the interior is contained: $$A^\circ\subseteq A$$ and that the interior is open: $$A^\circ\in\mathcal{N}(z)\text{ for all }z\in A^\circ$$

Answer 1

Let $z\in A^{\large\circ}$.

By definition, that means $A\in \mathcal{N}(z)$.

By the first property of neighborhood systems $z\in A$. This proves the first statement.

By the second property of neighbourhood systems, there is a $C\in \mathcal{N}(z)$ such that $A\in\mathcal{N}(c)$ for all $c\in C$. That means $c\in A^{\large\circ}$ for all $c\in C$, hence $C\subset A^{\large\circ}$. Since $\mathcal{N}(z)$ is a filter, $C\in \mathcal{N}(z)$ and $C\subset A^{\large\circ}$ imply $A^{\large\circ} \in \mathcal{N}(z)$. This proves the second statement.