Neighborhood Condition for Coarser Topology

Theorem
Let $S$ be a set.

Let $\tau_1$ and $\tau_2$ be two topologies on $S$.

Suppose that for all $z \in S$ and for all open neighborhoods $N_z$ of $z$ with respect to $\tau_1$, there exists $U \in \tau_2$ such that $U \subseteq N_z$.

Then $\tau_1$ is coarser than $\tau_2$.

Proof
Let $V \in \tau_1$.

For all $z \in V$, we have that $V$ is an open neighborhood of $z$ with respect to $\tau_1$.

Then for all $z \in V$, we can find $U_z \in \tau_2$ such that $z \in U_z \subseteq V$.

Then:


 * $\displaystyle V = \bigcup_{z \in V } \left\{ {z}\right\} \subseteq \bigcup_{z \in V } U_z \subseteq V$

By definition of set equality, it follows that:
 * $\displaystyle V = \bigcup_{z \in V } U_z \in \tau_2$

Hence, $\tau_1 \subseteq \tau_2$, which by definition means that $\tau_1$ is coarser than $\tau_2$.